EE101: RC and RL Circuits (with DC sources) M. B. Patil [email protected] www.ee.iitb.ac.in/~sequel Department of Electrical Engineering Indian Institute of Technology Bombay M. B. Patil, IIT Bombay
EE101: RC and RL Circuits (with DC sources)
M. B. [email protected]
www.ee.iitb.ac.in/~sequel
Department of Electrical EngineeringIndian Institute of Technology Bombay
M. B. Patil, IIT Bombay
Capacitors
v
i
Q
Q
insulator
i
conductor
conductor
Unit: Farad (F)
C =ǫ A
tt
* In practice, capacitors are available in a wide range of shapes and values, andthey differ significantly in the way they are fabricated.(http://en.wikipedia.org/wiki/Capacitor)
* To make C larger, we need (a) high ε, (b) large area, (c) small thickness.
* For a constant capacitance,
Q(t) = C v(t) ,dQ
dt= C
dv
dt, i.e, i(t) = C
dv
dt.
* If v = constant, i = 0, i.e., a capacitor behaves like an open circuit in DCconditions as one would expect from two conducting plates separated by aninsulator.
M. B. Patil, IIT Bombay
Capacitors
v
i
Q
Q
insulator
i
conductor
conductor
Unit: Farad (F)
C =ǫ A
tt
* In practice, capacitors are available in a wide range of shapes and values, andthey differ significantly in the way they are fabricated.(http://en.wikipedia.org/wiki/Capacitor)
* To make C larger, we need (a) high ε, (b) large area, (c) small thickness.
* For a constant capacitance,
Q(t) = C v(t) ,dQ
dt= C
dv
dt, i.e, i(t) = C
dv
dt.
* If v = constant, i = 0, i.e., a capacitor behaves like an open circuit in DCconditions as one would expect from two conducting plates separated by aninsulator.
M. B. Patil, IIT Bombay
Capacitors
v
i
Q
Q
insulator
i
conductor
conductor
Unit: Farad (F)
C =ǫ A
tt
* In practice, capacitors are available in a wide range of shapes and values, andthey differ significantly in the way they are fabricated.(http://en.wikipedia.org/wiki/Capacitor)
* To make C larger, we need (a) high ε, (b) large area, (c) small thickness.
* For a constant capacitance,
Q(t) = C v(t) ,dQ
dt= C
dv
dt, i.e, i(t) = C
dv
dt.
* If v = constant, i = 0, i.e., a capacitor behaves like an open circuit in DCconditions as one would expect from two conducting plates separated by aninsulator.
M. B. Patil, IIT Bombay
Capacitors
v
i
Q
Q
insulator
i
conductor
conductor
Unit: Farad (F)
C =ǫ A
tt
* In practice, capacitors are available in a wide range of shapes and values, andthey differ significantly in the way they are fabricated.(http://en.wikipedia.org/wiki/Capacitor)
* To make C larger, we need (a) high ε, (b) large area, (c) small thickness.
* For a constant capacitance,
Q(t) = C v(t) ,dQ
dt= C
dv
dt, i.e, i(t) = C
dv
dt.
* If v = constant, i = 0, i.e., a capacitor behaves like an open circuit in DCconditions as one would expect from two conducting plates separated by aninsulator.
M. B. Patil, IIT Bombay
Capacitors
v
i
Q
Q
insulator
i
conductor
conductor
Unit: Farad (F)
C =ǫ A
tt
* In practice, capacitors are available in a wide range of shapes and values, andthey differ significantly in the way they are fabricated.(http://en.wikipedia.org/wiki/Capacitor)
* To make C larger, we need (a) high ε, (b) large area, (c) small thickness.
* For a constant capacitance,
Q(t) = C v(t) ,dQ
dt= C
dv
dt, i.e, i(t) = C
dv
dt.
* If v = constant, i = 0, i.e., a capacitor behaves like an open circuit in DCconditions as one would expect from two conducting plates separated by aninsulator.
M. B. Patil, IIT Bombay
Example
i
v
0
20
−20
i (m
A)
for the given source current.
Plot v, p, and W versus time
Assume v(0) = 0 V, C= 5 mF.
i(t) = Cdv
dt
v(t) =1
C
∫i(t) dt
8
4
0
−4
v (V
)
p(t) = v(t)× i(t)
0.2
0.1
0
−0.1
−0.2
pow
er (
Wat
ts)
W(t) =∫
p(t) dt
0.2
0.1
0ener
gy (
J)
time (sec)
0 1 2 3 4 5 6
W(t) =∫
p(t) dt
= C∫
vdv
dtdt
= C∫
v dv
=1
2C v2
M. B. Patil, IIT Bombay
Example
i
v
0
20
−20
i (m
A)
for the given source current.
Plot v, p, and W versus time
Assume v(0) = 0 V, C= 5 mF.
i(t) = Cdv
dt
v(t) =1
C
∫i(t) dt
8
4
0
−4
v (V
)
p(t) = v(t)× i(t)
0.2
0.1
0
−0.1
−0.2
pow
er (
Wat
ts)
W(t) =∫
p(t) dt
0.2
0.1
0ener
gy (
J)
time (sec)
0 1 2 3 4 5 6
W(t) =∫
p(t) dt
= C∫
vdv
dtdt
= C∫
v dv
=1
2C v2
M. B. Patil, IIT Bombay
Example
i
v
0
20
−20
i (m
A)
for the given source current.
Plot v, p, and W versus time
Assume v(0) = 0 V, C= 5 mF.
i(t) = Cdv
dt
v(t) =1
C
∫i(t) dt
8
4
0
−4
v (V
)
p(t) = v(t)× i(t)
0.2
0.1
0
−0.1
−0.2
pow
er (
Wat
ts)
W(t) =∫
p(t) dt
0.2
0.1
0ener
gy (
J)
time (sec)
0 1 2 3 4 5 6
W(t) =∫
p(t) dt
= C∫
vdv
dtdt
= C∫
v dv
=1
2C v2
M. B. Patil, IIT Bombay
Example
i
v
0
20
−20
i (m
A)
for the given source current.
Plot v, p, and W versus time
Assume v(0) = 0 V, C= 5 mF.
i(t) = Cdv
dt
v(t) =1
C
∫i(t) dt
8
4
0
−4
v (V
)
p(t) = v(t)× i(t)
0.2
0.1
0
−0.1
−0.2
pow
er (
Wat
ts)
W(t) =∫
p(t) dt
0.2
0.1
0ener
gy (
J)
time (sec)
0 1 2 3 4 5 6
W(t) =∫
p(t) dt
= C∫
vdv
dtdt
= C∫
v dv
=1
2C v2
M. B. Patil, IIT Bombay
Example
i
v
0
20
−20
i (m
A)
for the given source current.
Plot v, p, and W versus time
Assume v(0) = 0 V, C= 5 mF.
i(t) = Cdv
dt
v(t) =1
C
∫i(t) dt
8
4
0
−4
v (V
)
p(t) = v(t)× i(t)
0.2
0.1
0
−0.1
−0.2
pow
er (
Wat
ts)
W(t) =∫
p(t) dt
0.2
0.1
0ener
gy (
J)
time (sec)
0 1 2 3 4 5 6
W(t) =∫
p(t) dt
= C∫
vdv
dtdt
= C∫
v dv
=1
2C v2
M. B. Patil, IIT Bombay
Example
i
v
0
20
−20
i (m
A)
for the given source current.
Plot v, p, and W versus time
Assume v(0) = 0 V, C= 5 mF.
i(t) = Cdv
dt
v(t) =1
C
∫i(t) dt
8
4
0
−4
v (V
)
p(t) = v(t)× i(t)
0.2
0.1
0
−0.1
−0.2
pow
er (
Wat
ts)
W(t) =∫
p(t) dt
0.2
0.1
0ener
gy (
J)
time (sec)
0 1 2 3 4 5 6
W(t) =∫
p(t) dt
= C∫
vdv
dtdt
= C∫
v dv
=1
2C v2
M. B. Patil, IIT Bombay
Example
i
v
0
20
−20
i (m
A)
for the given source current.
Plot v, p, and W versus time
Assume v(0) = 0 V, C= 5 mF.
i(t) = Cdv
dt
v(t) =1
C
∫i(t) dt
8
4
0
−4
v (V
)
p(t) = v(t)× i(t)
0.2
0.1
0
−0.1
−0.2
pow
er (
Wat
ts)
W(t) =∫
p(t) dt
0.2
0.1
0ener
gy (
J)
time (sec)
0 1 2 3 4 5 6
W(t) =∫
p(t) dt
= C∫
vdv
dtdt
= C∫
v dv
=1
2C v2
M. B. Patil, IIT Bombay
Example
i
v
0
20
−20
i (m
A)
for the given source current.
Plot v, p, and W versus time
Assume v(0) = 0 V, C= 5 mF.
i(t) = Cdv
dt
v(t) =1
C
∫i(t) dt
8
4
0
−4
v (V
)
p(t) = v(t)× i(t)
0.2
0.1
0
−0.1
−0.2
pow
er (
Wat
ts)
W(t) =∫
p(t) dt
0.2
0.1
0ener
gy (
J)
time (sec)
0 1 2 3 4 5 6
W(t) =∫
p(t) dt
= C∫
vdv
dtdt
= C∫
v dv
=1
2C v2
M. B. Patil, IIT Bombay
Home work
v
i i (mA)
0 1 2
20
time (sec)
* For the given source current, plot v(t), p(t), and W (t), assuming v(0) = 0 V ,C = 5 mF .
* Verify your results with circuit simulation.
M. B. Patil, IIT Bombay
Home work
v
i i (mA)
0 1 2
20
time (sec)
* For the given source current, plot v(t), p(t), and W (t), assuming v(0) = 0 V ,C = 5 mF .
* Verify your results with circuit simulation.
M. B. Patil, IIT Bombay
Home work
v
i i (mA)
0 1 2
20
time (sec)
* For the given source current, plot v(t), p(t), and W (t), assuming v(0) = 0 V ,C = 5 mF .
* Verify your results with circuit simulation.
M. B. Patil, IIT Bombay
Inductors
core Magnetic field lines
v
i L
Symbol
Units: Henry (H)
* An inductor is basically a conducting coil wound around a “core.”
* V = Ndφ
dt= N
d
dt(B · A) = N
d
dt
»„µN i
l
«A
–.
Compare with v = Ldi
dt.
⇒ L = µN2 A
l= µrµ0 N2 A
l.
* To make L larger, we need (a) high µr , (b) large area, (c) large number of turns.
* For 99.8 % pure iron, µr ' 5, 000 .For “supermalloy” (Ni: 79 %, Mo: 5 %, Fe): µr ' 106 .
* If i = constant, v = 0, i.e., an inductor behaves like a short circuit in DCconditions as one would expect from a highly conducting coil.
* Note: B = µH is an approximation. In practice, B may be a nonlinear functionof H, depending on the core material.
M. B. Patil, IIT Bombay
Inductors
core Magnetic field lines
v
i L
Symbol
Units: Henry (H)
* An inductor is basically a conducting coil wound around a “core.”
* V = Ndφ
dt= N
d
dt(B · A) = N
d
dt
»„µN i
l
«A
–.
Compare with v = Ldi
dt.
⇒ L = µN2 A
l= µrµ0 N2 A
l.
* To make L larger, we need (a) high µr , (b) large area, (c) large number of turns.
* For 99.8 % pure iron, µr ' 5, 000 .For “supermalloy” (Ni: 79 %, Mo: 5 %, Fe): µr ' 106 .
* If i = constant, v = 0, i.e., an inductor behaves like a short circuit in DCconditions as one would expect from a highly conducting coil.
* Note: B = µH is an approximation. In practice, B may be a nonlinear functionof H, depending on the core material.
M. B. Patil, IIT Bombay
Inductors
core Magnetic field lines
v
i L
Symbol
Units: Henry (H)
* An inductor is basically a conducting coil wound around a “core.”
* V = Ndφ
dt= N
d
dt(B · A) = N
d
dt
»„µN i
l
«A
–.
Compare with v = Ldi
dt.
⇒ L = µN2 A
l= µrµ0 N2 A
l.
* To make L larger, we need (a) high µr , (b) large area, (c) large number of turns.
* For 99.8 % pure iron, µr ' 5, 000 .For “supermalloy” (Ni: 79 %, Mo: 5 %, Fe): µr ' 106 .
* If i = constant, v = 0, i.e., an inductor behaves like a short circuit in DCconditions as one would expect from a highly conducting coil.
* Note: B = µH is an approximation. In practice, B may be a nonlinear functionof H, depending on the core material.
M. B. Patil, IIT Bombay
Inductors
core Magnetic field lines
v
i L
Symbol
Units: Henry (H)
* An inductor is basically a conducting coil wound around a “core.”
* V = Ndφ
dt= N
d
dt(B · A) = N
d
dt
»„µN i
l
«A
–.
Compare with v = Ldi
dt.
⇒ L = µN2 A
l= µrµ0 N2 A
l.
* To make L larger, we need (a) high µr , (b) large area, (c) large number of turns.
* For 99.8 % pure iron, µr ' 5, 000 .For “supermalloy” (Ni: 79 %, Mo: 5 %, Fe): µr ' 106 .
* If i = constant, v = 0, i.e., an inductor behaves like a short circuit in DCconditions as one would expect from a highly conducting coil.
* Note: B = µH is an approximation. In practice, B may be a nonlinear functionof H, depending on the core material.
M. B. Patil, IIT Bombay
Inductors
core Magnetic field lines
v
i L
Symbol
Units: Henry (H)
* An inductor is basically a conducting coil wound around a “core.”
* V = Ndφ
dt= N
d
dt(B · A) = N
d
dt
»„µN i
l
«A
–.
Compare with v = Ldi
dt.
⇒ L = µN2 A
l= µrµ0 N2 A
l.
* To make L larger, we need (a) high µr , (b) large area, (c) large number of turns.
* For 99.8 % pure iron, µr ' 5, 000 .For “supermalloy” (Ni: 79 %, Mo: 5 %, Fe): µr ' 106 .
* If i = constant, v = 0, i.e., an inductor behaves like a short circuit in DCconditions as one would expect from a highly conducting coil.
* Note: B = µH is an approximation. In practice, B may be a nonlinear functionof H, depending on the core material.
M. B. Patil, IIT Bombay
Inductors
core Magnetic field lines
v
i L
Symbol
Units: Henry (H)
* An inductor is basically a conducting coil wound around a “core.”
* V = Ndφ
dt= N
d
dt(B · A) = N
d
dt
»„µN i
l
«A
–.
Compare with v = Ldi
dt.
⇒ L = µN2 A
l= µrµ0 N2 A
l.
* To make L larger, we need (a) high µr , (b) large area, (c) large number of turns.
* For 99.8 % pure iron, µr ' 5, 000 .For “supermalloy” (Ni: 79 %, Mo: 5 %, Fe): µr ' 106 .
* If i = constant, v = 0, i.e., an inductor behaves like a short circuit in DCconditions as one would expect from a highly conducting coil.
* Note: B = µH is an approximation. In practice, B may be a nonlinear functionof H, depending on the core material.
M. B. Patil, IIT Bombay
Inductors
core Magnetic field lines
v
i L
Symbol
Units: Henry (H)
* An inductor is basically a conducting coil wound around a “core.”
* V = Ndφ
dt= N
d
dt(B · A) = N
d
dt
»„µN i
l
«A
–.
Compare with v = Ldi
dt.
⇒ L = µN2 A
l= µrµ0 N2 A
l.
* To make L larger, we need (a) high µr , (b) large area, (c) large number of turns.
* For 99.8 % pure iron, µr ' 5, 000 .For “supermalloy” (Ni: 79 %, Mo: 5 %, Fe): µr ' 106 .
* If i = constant, v = 0, i.e., an inductor behaves like a short circuit in DCconditions as one would expect from a highly conducting coil.
* Note: B = µH is an approximation. In practice, B may be a nonlinear functionof H, depending on the core material.
M. B. Patil, IIT Bombay
RC circuits with DC sources
i
Cv
A
B
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
Cv
A
B
≡ VTh
RTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v → VTh = RThCdv
dt+ v .
* Homogeneous solution:
dv
dt+
1
τv = 0 , where τ = RTh C is the “time constant.”
→ v (h) = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differntial equation. Weknow that all time derivatives will vanish as t →∞ , making i = 0, and we getv (p) = VTh as a particular solution (which happens to be simply a constant).
* v = v (h) + v (p) = K exp(−t/τ) + VTh .
* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained fromknown conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources
i
Cv
A
B
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
Cv
A
B
≡ VTh
RTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v → VTh = RThCdv
dt+ v .
* Homogeneous solution:
dv
dt+
1
τv = 0 , where τ = RTh C is the “time constant.”
→ v (h) = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differntial equation. Weknow that all time derivatives will vanish as t →∞ , making i = 0, and we getv (p) = VTh as a particular solution (which happens to be simply a constant).
* v = v (h) + v (p) = K exp(−t/τ) + VTh .
* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained fromknown conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources
i
Cv
A
B
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
Cv
A
B
≡ VTh
RTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v → VTh = RThCdv
dt+ v .
* Homogeneous solution:
dv
dt+
1
τv = 0 , where τ = RTh C is the “time constant.”
→ v (h) = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differntial equation. Weknow that all time derivatives will vanish as t →∞ , making i = 0, and we getv (p) = VTh as a particular solution (which happens to be simply a constant).
* v = v (h) + v (p) = K exp(−t/τ) + VTh .
* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained fromknown conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources
i
Cv
A
B
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
Cv
A
B
≡ VTh
RTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v → VTh = RThCdv
dt+ v .
* Homogeneous solution:
dv
dt+
1
τv = 0 , where τ = RTh C is the “time constant.”
→ v (h) = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differntial equation. Weknow that all time derivatives will vanish as t →∞ , making i = 0, and we getv (p) = VTh as a particular solution (which happens to be simply a constant).
* v = v (h) + v (p) = K exp(−t/τ) + VTh .
* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained fromknown conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources
i
Cv
A
B
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
Cv
A
B
≡ VTh
RTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v → VTh = RThCdv
dt+ v .
* Homogeneous solution:
dv
dt+
1
τv = 0 , where τ = RTh C is the “time constant.”
→ v (h) = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differntial equation. Weknow that all time derivatives will vanish as t →∞ , making i = 0, and we getv (p) = VTh as a particular solution (which happens to be simply a constant).
* v = v (h) + v (p) = K exp(−t/τ) + VTh .
* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained fromknown conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources
i
Cv
A
B
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
Cv
A
B
≡ VTh
RTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v → VTh = RThCdv
dt+ v .
* Homogeneous solution:
dv
dt+
1
τv = 0 , where τ = RTh C is the “time constant.”
→ v (h) = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differntial equation. Weknow that all time derivatives will vanish as t →∞ , making i = 0, and we getv (p) = VTh as a particular solution (which happens to be simply a constant).
* v = v (h) + v (p) = K exp(−t/τ) + VTh .
* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained fromknown conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources
i
Cv
A
B
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
Cv
A
B
≡ VTh
RTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v → VTh = RThCdv
dt+ v .
* Homogeneous solution:
dv
dt+
1
τv = 0 , where τ = RTh C is the “time constant.”
→ v (h) = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differntial equation. Weknow that all time derivatives will vanish as t →∞ , making i = 0, and we getv (p) = VTh as a particular solution (which happens to be simply a constant).
* v = v (h) + v (p) = K exp(−t/τ) + VTh .
* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained fromknown conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources
i
Cv
A
B
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
Cv
A
B
≡ VTh
RTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v → VTh = RThCdv
dt+ v .
* Homogeneous solution:
dv
dt+
1
τv = 0 , where τ = RTh C is the “time constant.”
→ v (h) = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differntial equation. Weknow that all time derivatives will vanish as t →∞ , making i = 0, and we getv (p) = VTh as a particular solution (which happens to be simply a constant).
* v = v (h) + v (p) = K exp(−t/τ) + VTh .
* In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained fromknown conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources (continued)
i i
Cv
Cv
A
B
A
B
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
≡ VTh
RTh
* If all sources are DC (constant), we havev(t) = A exp(−t/τ) + B , τ = RC .
* i(t) = Cdv
dt= C × A exp(−t/τ)
„− 1
τ
«≡ A′ exp(−t/τ) .
* As t →∞, i → 0, i.e., the capacitor behaves like an open circuit since allderivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) inthe circuit can be expressed asx(t) = K1 exp(−t/τ) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
RC circuits with DC sources (continued)
i i
Cv
Cv
A
B
A
B
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
≡ VTh
RTh
* If all sources are DC (constant), we havev(t) = A exp(−t/τ) + B , τ = RC .
* i(t) = Cdv
dt= C × A exp(−t/τ)
„− 1
τ
«≡ A′ exp(−t/τ) .
* As t →∞, i → 0, i.e., the capacitor behaves like an open circuit since allderivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) inthe circuit can be expressed asx(t) = K1 exp(−t/τ) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
RC circuits with DC sources (continued)
i i
Cv
Cv
A
B
A
B
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
≡ VTh
RTh
* If all sources are DC (constant), we havev(t) = A exp(−t/τ) + B , τ = RC .
* i(t) = Cdv
dt= C × A exp(−t/τ)
„− 1
τ
«≡ A′ exp(−t/τ) .
* As t →∞, i → 0, i.e., the capacitor behaves like an open circuit since allderivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) inthe circuit can be expressed asx(t) = K1 exp(−t/τ) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
RC circuits with DC sources (continued)
i i
Cv
Cv
A
B
A
B
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
≡ VTh
RTh
* If all sources are DC (constant), we havev(t) = A exp(−t/τ) + B , τ = RC .
* i(t) = Cdv
dt= C × A exp(−t/τ)
„− 1
τ
«≡ A′ exp(−t/τ) .
* As t →∞, i → 0, i.e., the capacitor behaves like an open circuit since allderivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) inthe circuit can be expressed asx(t) = K1 exp(−t/τ) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
RL circuits with DC sources
v
iCircuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
A
B
v
i
A
B
≡ VTh
RTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + Ldi
dt.
* Homogeneous solution:
di
dt+
1
τi = 0 , where τ = L/RTh
→ i (h) = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differntial equation. Weknow that all time derivatives will vanish as t →∞ , making v = 0, and we geti (p) = VTh/RTh as a particular solution (which happens to be simply a constant).
* i = i (h) + i (p) = K exp(−t/τ) + VTh/RTh .
* In general, i(t) = A exp(−t/τ) + B , where A and B can be obtained fromknown conditions on i .
M. B. Patil, IIT Bombay
RL circuits with DC sources
v
iCircuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
A
B
v
i
A
B
≡ VTh
RTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + Ldi
dt.
* Homogeneous solution:
di
dt+
1
τi = 0 , where τ = L/RTh
→ i (h) = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differntial equation. Weknow that all time derivatives will vanish as t →∞ , making v = 0, and we geti (p) = VTh/RTh as a particular solution (which happens to be simply a constant).
* i = i (h) + i (p) = K exp(−t/τ) + VTh/RTh .
* In general, i(t) = A exp(−t/τ) + B , where A and B can be obtained fromknown conditions on i .
M. B. Patil, IIT Bombay
RL circuits with DC sources
v
iCircuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
A
B
v
i
A
B
≡ VTh
RTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + Ldi
dt.
* Homogeneous solution:
di
dt+
1
τi = 0 , where τ = L/RTh
→ i (h) = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differntial equation. Weknow that all time derivatives will vanish as t →∞ , making v = 0, and we geti (p) = VTh/RTh as a particular solution (which happens to be simply a constant).
* i = i (h) + i (p) = K exp(−t/τ) + VTh/RTh .
* In general, i(t) = A exp(−t/τ) + B , where A and B can be obtained fromknown conditions on i .
M. B. Patil, IIT Bombay
RL circuits with DC sources
v
iCircuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
A
B
v
i
A
B
≡ VTh
RTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + Ldi
dt.
* Homogeneous solution:
di
dt+
1
τi = 0 , where τ = L/RTh
→ i (h) = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differntial equation. Weknow that all time derivatives will vanish as t →∞ , making v = 0, and we geti (p) = VTh/RTh as a particular solution (which happens to be simply a constant).
* i = i (h) + i (p) = K exp(−t/τ) + VTh/RTh .
* In general, i(t) = A exp(−t/τ) + B , where A and B can be obtained fromknown conditions on i .
M. B. Patil, IIT Bombay
RL circuits with DC sources
v
iCircuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
A
B
v
i
A
B
≡ VTh
RTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + Ldi
dt.
* Homogeneous solution:
di
dt+
1
τi = 0 , where τ = L/RTh
→ i (h) = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differntial equation. Weknow that all time derivatives will vanish as t →∞ , making v = 0, and we geti (p) = VTh/RTh as a particular solution (which happens to be simply a constant).
* i = i (h) + i (p) = K exp(−t/τ) + VTh/RTh .
* In general, i(t) = A exp(−t/τ) + B , where A and B can be obtained fromknown conditions on i .
M. B. Patil, IIT Bombay
RL circuits with DC sources
v
iCircuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
A
B
v
i
A
B
≡ VTh
RTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + Ldi
dt.
* Homogeneous solution:
di
dt+
1
τi = 0 , where τ = L/RTh
→ i (h) = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differntial equation. Weknow that all time derivatives will vanish as t →∞ , making v = 0, and we geti (p) = VTh/RTh as a particular solution (which happens to be simply a constant).
* i = i (h) + i (p) = K exp(−t/τ) + VTh/RTh .
* In general, i(t) = A exp(−t/τ) + B , where A and B can be obtained fromknown conditions on i .
M. B. Patil, IIT Bombay
RL circuits with DC sources
v
iCircuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
A
B
v
i
A
B
≡ VTh
RTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + Ldi
dt.
* Homogeneous solution:
di
dt+
1
τi = 0 , where τ = L/RTh
→ i (h) = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differntial equation. Weknow that all time derivatives will vanish as t →∞ , making v = 0, and we geti (p) = VTh/RTh as a particular solution (which happens to be simply a constant).
* i = i (h) + i (p) = K exp(−t/τ) + VTh/RTh .
* In general, i(t) = A exp(−t/τ) + B , where A and B can be obtained fromknown conditions on i .
M. B. Patil, IIT Bombay
RL circuits with DC sources
v
iCircuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
A
B
v
i
A
B
≡ VTh
RTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + Ldi
dt.
* Homogeneous solution:
di
dt+
1
τi = 0 , where τ = L/RTh
→ i (h) = K exp(−t/τ) .
* Particular solution is a specific function that satisfies the differntial equation. Weknow that all time derivatives will vanish as t →∞ , making v = 0, and we geti (p) = VTh/RTh as a particular solution (which happens to be simply a constant).
* i = i (h) + i (p) = K exp(−t/τ) + VTh/RTh .
* In general, i(t) = A exp(−t/τ) + B , where A and B can be obtained fromknown conditions on i .
M. B. Patil, IIT Bombay
RL circuits with DC sources (continued)
vv
i iCircuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
A
B
A
B
≡ VTh
RTh
* If all sources are DC (constant), we havei(t) = A exp(−t/τ) + B , τ = L/R .
* v(t) = Ldi
dt= L× A exp(−t/τ)
„− 1
τ
«≡ A′ exp(−t/τ) .
* As t →∞, v → 0, i.e., the inductor behaves like a short circuit since allderivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) inthe circuit can be expressed asx(t) = K1 exp(−t/τ) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
RL circuits with DC sources (continued)
vv
i iCircuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
A
B
A
B
≡ VTh
RTh
* If all sources are DC (constant), we havei(t) = A exp(−t/τ) + B , τ = L/R .
* v(t) = Ldi
dt= L× A exp(−t/τ)
„− 1
τ
«≡ A′ exp(−t/τ) .
* As t →∞, v → 0, i.e., the inductor behaves like a short circuit since allderivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) inthe circuit can be expressed asx(t) = K1 exp(−t/τ) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
RL circuits with DC sources (continued)
vv
i iCircuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
A
B
A
B
≡ VTh
RTh
* If all sources are DC (constant), we havei(t) = A exp(−t/τ) + B , τ = L/R .
* v(t) = Ldi
dt= L× A exp(−t/τ)
„− 1
τ
«≡ A′ exp(−t/τ) .
* As t →∞, v → 0, i.e., the inductor behaves like a short circuit since allderivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) inthe circuit can be expressed asx(t) = K1 exp(−t/τ) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
RL circuits with DC sources (continued)
vv
i iCircuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
A
B
A
B
≡ VTh
RTh
* If all sources are DC (constant), we havei(t) = A exp(−t/τ) + B , τ = L/R .
* v(t) = Ldi
dt= L× A exp(−t/τ)
„− 1
τ
«≡ A′ exp(−t/τ) .
* As t →∞, v → 0, i.e., the inductor behaves like a short circuit since allderivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) inthe circuit can be expressed asx(t) = K1 exp(−t/τ) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
RC circuits: Can Vc change “suddenly?”
i
t0 V
5 V
C = 1 µFVc
Vs
Vs
R = 1 k
Vc(0)=0 V
* Vs changes from 0 V (at t = 0−), to 5 V (at t = 0+). As a result of thischange, Vc will rise. How fast can Vc change?
* For example, what would happen if Vc changes by 1 V in 1 µs at a constantrate of 1 V /1 µs = 106 V /s?
* i = CdVc
dt= 1 µF × 106 V
s= 1 A .
* With i = 1 A, the voltage drop across R would be 1000 V ! Not allowed by KVL.
* We conclude that Vc (0+) = Vc (0−)⇒ A capacitor does not allow abruptchanges in Vc if there is a finite resistance in the circuit.
* Similarly, an inductor does not allow abrupt changes in iL.
M. B. Patil, IIT Bombay
RC circuits: Can Vc change “suddenly?”
i
t0 V
5 V
C = 1 µFVc
Vs
Vs
R = 1 k
Vc(0)=0 V
* Vs changes from 0 V (at t = 0−), to 5 V (at t = 0+). As a result of thischange, Vc will rise. How fast can Vc change?
* For example, what would happen if Vc changes by 1 V in 1 µs at a constantrate of 1 V /1 µs = 106 V /s?
* i = CdVc
dt= 1 µF × 106 V
s= 1 A .
* With i = 1 A, the voltage drop across R would be 1000 V ! Not allowed by KVL.
* We conclude that Vc (0+) = Vc (0−)⇒ A capacitor does not allow abruptchanges in Vc if there is a finite resistance in the circuit.
* Similarly, an inductor does not allow abrupt changes in iL.
M. B. Patil, IIT Bombay
RC circuits: Can Vc change “suddenly?”
i
t0 V
5 V
C = 1 µFVc
Vs
Vs
R = 1 k
Vc(0)=0 V
* Vs changes from 0 V (at t = 0−), to 5 V (at t = 0+). As a result of thischange, Vc will rise. How fast can Vc change?
* For example, what would happen if Vc changes by 1 V in 1 µs at a constantrate of 1 V /1 µs = 106 V /s?
* i = CdVc
dt= 1 µF × 106 V
s= 1 A .
* With i = 1 A, the voltage drop across R would be 1000 V ! Not allowed by KVL.
* We conclude that Vc (0+) = Vc (0−)⇒ A capacitor does not allow abruptchanges in Vc if there is a finite resistance in the circuit.
* Similarly, an inductor does not allow abrupt changes in iL.
M. B. Patil, IIT Bombay
RC circuits: Can Vc change “suddenly?”
i
t0 V
5 V
C = 1 µFVc
Vs
Vs
R = 1 k
Vc(0)=0 V
* Vs changes from 0 V (at t = 0−), to 5 V (at t = 0+). As a result of thischange, Vc will rise. How fast can Vc change?
* For example, what would happen if Vc changes by 1 V in 1 µs at a constantrate of 1 V /1 µs = 106 V /s?
* i = CdVc
dt= 1 µF × 106 V
s= 1 A .
* With i = 1 A, the voltage drop across R would be 1000 V ! Not allowed by KVL.
* We conclude that Vc (0+) = Vc (0−)⇒ A capacitor does not allow abruptchanges in Vc if there is a finite resistance in the circuit.
* Similarly, an inductor does not allow abrupt changes in iL.
M. B. Patil, IIT Bombay
RC circuits: Can Vc change “suddenly?”
i
t0 V
5 V
C = 1 µFVc
Vs
Vs
R = 1 k
Vc(0)=0 V
* Vs changes from 0 V (at t = 0−), to 5 V (at t = 0+). As a result of thischange, Vc will rise. How fast can Vc change?
* For example, what would happen if Vc changes by 1 V in 1 µs at a constantrate of 1 V /1 µs = 106 V /s?
* i = CdVc
dt= 1 µF × 106 V
s= 1 A .
* With i = 1 A, the voltage drop across R would be 1000 V ! Not allowed by KVL.
* We conclude that Vc (0+) = Vc (0−)⇒ A capacitor does not allow abruptchanges in Vc if there is a finite resistance in the circuit.
* Similarly, an inductor does not allow abrupt changes in iL.
M. B. Patil, IIT Bombay
RC circuits: Can Vc change “suddenly?”
i
t0 V
5 V
C = 1 µFVc
Vs
Vs
R = 1 k
Vc(0)=0 V
* Vs changes from 0 V (at t = 0−), to 5 V (at t = 0+). As a result of thischange, Vc will rise. How fast can Vc change?
* For example, what would happen if Vc changes by 1 V in 1 µs at a constantrate of 1 V /1 µs = 106 V /s?
* i = CdVc
dt= 1 µF × 106 V
s= 1 A .
* With i = 1 A, the voltage drop across R would be 1000 V ! Not allowed by KVL.
* We conclude that Vc (0+) = Vc (0−)⇒ A capacitor does not allow abruptchanges in Vc if there is a finite resistance in the circuit.
* Similarly, an inductor does not allow abrupt changes in iL.
M. B. Patil, IIT Bombay
RC circuits: Can Vc change “suddenly?”
i
t0 V
5 V
C = 1 µFVc
Vs
Vs
R = 1 k
Vc(0)=0 V
* Vs changes from 0 V (at t = 0−), to 5 V (at t = 0+). As a result of thischange, Vc will rise. How fast can Vc change?
* For example, what would happen if Vc changes by 1 V in 1 µs at a constantrate of 1 V /1 µs = 106 V /s?
* i = CdVc
dt= 1 µF × 106 V
s= 1 A .
* With i = 1 A, the voltage drop across R would be 1000 V ! Not allowed by KVL.
* We conclude that Vc (0+) = Vc (0−)⇒ A capacitor does not allow abruptchanges in Vc if there is a finite resistance in the circuit.
* Similarly, an inductor does not allow abrupt changes in iL.
M. B. Patil, IIT Bombay
Analysis of RC/RL circuits with a piece-wise constant source
* Identify intervals in which the source voltages/currents are constant.For example,
(1) t < t1
(3) t > t2
(2) t1 < t < t2
0 t2t1
Vs
* For any current or voltage x(t), write general expressions such as,x(t) = A1 exp(−t/τ) + B1 , t < t1 ,x(t) = A2 exp(−t/τ) + B2 , t1 < t < t2 ,x(t) = A3 exp(−t/τ) + B3 , t > t2 .
* Work out suitable conditions on x(t) at specific time points using
(a) If the source voltage/current has not changed for a “long” time(long compared to τ), all derivatives are zero.
⇒ iC = CdVc
dt= 0 , and VL = L
diL
dt= 0 .
(b) When a source voltage (or current) changes, say, at t = t0 ,Vc (t) or iL(t) cannot change abruptly, i.e.,
Vc (t+0 ) = Vc (t−0 ) , and iL(t+
0 ) = iL(t−0 ) .
* Compute A1, B1, · · · using the conditions on x(t).
M. B. Patil, IIT Bombay
Analysis of RC/RL circuits with a piece-wise constant source
* Identify intervals in which the source voltages/currents are constant.For example,
(1) t < t1
(3) t > t2
(2) t1 < t < t2
0 t2t1
Vs
* For any current or voltage x(t), write general expressions such as,x(t) = A1 exp(−t/τ) + B1 , t < t1 ,x(t) = A2 exp(−t/τ) + B2 , t1 < t < t2 ,x(t) = A3 exp(−t/τ) + B3 , t > t2 .
* Work out suitable conditions on x(t) at specific time points using
(a) If the source voltage/current has not changed for a “long” time(long compared to τ), all derivatives are zero.
⇒ iC = CdVc
dt= 0 , and VL = L
diL
dt= 0 .
(b) When a source voltage (or current) changes, say, at t = t0 ,Vc (t) or iL(t) cannot change abruptly, i.e.,
Vc (t+0 ) = Vc (t−0 ) , and iL(t+
0 ) = iL(t−0 ) .
* Compute A1, B1, · · · using the conditions on x(t).
M. B. Patil, IIT Bombay
Analysis of RC/RL circuits with a piece-wise constant source
* Identify intervals in which the source voltages/currents are constant.For example,
(1) t < t1
(3) t > t2
(2) t1 < t < t2
0 t2t1
Vs
* For any current or voltage x(t), write general expressions such as,x(t) = A1 exp(−t/τ) + B1 , t < t1 ,x(t) = A2 exp(−t/τ) + B2 , t1 < t < t2 ,x(t) = A3 exp(−t/τ) + B3 , t > t2 .
* Work out suitable conditions on x(t) at specific time points using
(a) If the source voltage/current has not changed for a “long” time(long compared to τ), all derivatives are zero.
⇒ iC = CdVc
dt= 0 , and VL = L
diL
dt= 0 .
(b) When a source voltage (or current) changes, say, at t = t0 ,Vc (t) or iL(t) cannot change abruptly, i.e.,
Vc (t+0 ) = Vc (t−0 ) , and iL(t+
0 ) = iL(t−0 ) .
* Compute A1, B1, · · · using the conditions on x(t).
M. B. Patil, IIT Bombay
Analysis of RC/RL circuits with a piece-wise constant source
* Identify intervals in which the source voltages/currents are constant.For example,
(1) t < t1
(3) t > t2
(2) t1 < t < t2
0 t2t1
Vs
* For any current or voltage x(t), write general expressions such as,x(t) = A1 exp(−t/τ) + B1 , t < t1 ,x(t) = A2 exp(−t/τ) + B2 , t1 < t < t2 ,x(t) = A3 exp(−t/τ) + B3 , t > t2 .
* Work out suitable conditions on x(t) at specific time points using
(a) If the source voltage/current has not changed for a “long” time(long compared to τ), all derivatives are zero.
⇒ iC = CdVc
dt= 0 , and VL = L
diL
dt= 0 .
(b) When a source voltage (or current) changes, say, at t = t0 ,Vc (t) or iL(t) cannot change abruptly, i.e.,
Vc (t+0 ) = Vc (t−0 ) , and iL(t+
0 ) = iL(t−0 ) .
* Compute A1, B1, · · · using the conditions on x(t).
M. B. Patil, IIT Bombay
Analysis of RC/RL circuits with a piece-wise constant source
* Identify intervals in which the source voltages/currents are constant.For example,
(1) t < t1
(3) t > t2
(2) t1 < t < t2
0 t2t1
Vs
* For any current or voltage x(t), write general expressions such as,x(t) = A1 exp(−t/τ) + B1 , t < t1 ,x(t) = A2 exp(−t/τ) + B2 , t1 < t < t2 ,x(t) = A3 exp(−t/τ) + B3 , t > t2 .
* Work out suitable conditions on x(t) at specific time points using
(a) If the source voltage/current has not changed for a “long” time(long compared to τ), all derivatives are zero.
⇒ iC = CdVc
dt= 0 , and VL = L
diL
dt= 0 .
(b) When a source voltage (or current) changes, say, at t = t0 ,Vc (t) or iL(t) cannot change abruptly, i.e.,
Vc (t+0 ) = Vc (t−0 ) , and iL(t+
0 ) = iL(t−0 ) .
* Compute A1, B1, · · · using the conditions on x(t).
M. B. Patil, IIT Bombay
Analysis of RC/RL circuits with a piece-wise constant source
* Identify intervals in which the source voltages/currents are constant.For example,
(1) t < t1
(3) t > t2
(2) t1 < t < t2
0 t2t1
Vs
* For any current or voltage x(t), write general expressions such as,x(t) = A1 exp(−t/τ) + B1 , t < t1 ,x(t) = A2 exp(−t/τ) + B2 , t1 < t < t2 ,x(t) = A3 exp(−t/τ) + B3 , t > t2 .
* Work out suitable conditions on x(t) at specific time points using
(a) If the source voltage/current has not changed for a “long” time(long compared to τ), all derivatives are zero.
⇒ iC = CdVc
dt= 0 , and VL = L
diL
dt= 0 .
(b) When a source voltage (or current) changes, say, at t = t0 ,Vc (t) or iL(t) cannot change abruptly, i.e.,
Vc (t+0 ) = Vc (t−0 ) , and iL(t+
0 ) = iL(t−0 ) .
* Compute A1, B1, · · · using the conditions on x(t).
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
i
t0 V
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0−) = Vs(0−) = 0 V
v(0+) ≃ v(0−) = 0 V
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
As t→∞ , i→ 0 → v(∞) = Vs(∞) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., A = V0 , B = −V0
t = 0+: 0 = A + B ,
t→∞: V0 = B .
v(t) = V0 [1− exp(−t/τ)]
i
t0 V
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
v(0−) = Vs(0−) = V0
v(0+) ≃ v(0−) = V0
As t→∞ , i→ 0 → v(∞) = Vs(∞) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A + B ,
i.e., A = V0 , B = 0
t→∞: 0 = B .
v(t) = V0 exp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
i
t0 V
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0−) = Vs(0−) = 0 V
v(0+) ≃ v(0−) = 0 V
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
As t→∞ , i→ 0 → v(∞) = Vs(∞) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., A = V0 , B = −V0
t = 0+: 0 = A + B ,
t→∞: V0 = B .
v(t) = V0 [1− exp(−t/τ)]
i
t0 V
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
v(0−) = Vs(0−) = V0
v(0+) ≃ v(0−) = V0
As t→∞ , i→ 0 → v(∞) = Vs(∞) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A + B ,
i.e., A = V0 , B = 0
t→∞: 0 = B .
v(t) = V0 exp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
i
t0 V
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0−) = Vs(0−) = 0 V
v(0+) ≃ v(0−) = 0 V
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
As t→∞ , i→ 0 → v(∞) = Vs(∞) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., A = V0 , B = −V0
t = 0+: 0 = A + B ,
t→∞: V0 = B .
v(t) = V0 [1− exp(−t/τ)]
i
t0 V
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
v(0−) = Vs(0−) = V0
v(0+) ≃ v(0−) = V0
As t→∞ , i→ 0 → v(∞) = Vs(∞) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A + B ,
i.e., A = V0 , B = 0
t→∞: 0 = B .
v(t) = V0 exp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
i
t0 V
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0−) = Vs(0−) = 0 V
v(0+) ≃ v(0−) = 0 V
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
As t→∞ , i→ 0 → v(∞) = Vs(∞) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., A = V0 , B = −V0
t = 0+: 0 = A + B ,
t→∞: V0 = B .
v(t) = V0 [1− exp(−t/τ)]
i
t0 V
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
v(0−) = Vs(0−) = V0
v(0+) ≃ v(0−) = V0
As t→∞ , i→ 0 → v(∞) = Vs(∞) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A + B ,
i.e., A = V0 , B = 0
t→∞: 0 = B .
v(t) = V0 exp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
i
t0 V
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0−) = Vs(0−) = 0 V
v(0+) ≃ v(0−) = 0 V
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
As t→∞ , i→ 0 → v(∞) = Vs(∞) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., A = V0 , B = −V0
t = 0+: 0 = A + B ,
t→∞: V0 = B .
v(t) = V0 [1− exp(−t/τ)]
i
t0 V
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
v(0−) = Vs(0−) = V0
v(0+) ≃ v(0−) = V0
As t→∞ , i→ 0 → v(∞) = Vs(∞) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A + B ,
i.e., A = V0 , B = 0
t→∞: 0 = B .
v(t) = V0 exp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
i
t0 V
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0−) = Vs(0−) = 0 V
v(0+) ≃ v(0−) = 0 V
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
As t→∞ , i→ 0 → v(∞) = Vs(∞) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., A = V0 , B = −V0
t = 0+: 0 = A + B ,
t→∞: V0 = B .
v(t) = V0 [1− exp(−t/τ)]
i
t0 V
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
v(0−) = Vs(0−) = V0
v(0+) ≃ v(0−) = V0
As t→∞ , i→ 0 → v(∞) = Vs(∞) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A + B ,
i.e., A = V0 , B = 0
t→∞: 0 = B .
v(t) = V0 exp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
i
t0 V
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0−) = Vs(0−) = 0 V
v(0+) ≃ v(0−) = 0 V
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
As t→∞ , i→ 0 → v(∞) = Vs(∞) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., A = V0 , B = −V0
t = 0+: 0 = A + B ,
t→∞: V0 = B .
v(t) = V0 [1− exp(−t/τ)]
i
t0 V
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
v(0−) = Vs(0−) = V0
v(0+) ≃ v(0−) = V0
As t→∞ , i→ 0 → v(∞) = Vs(∞) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A + B ,
i.e., A = V0 , B = 0
t→∞: 0 = B .
v(t) = V0 exp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
i
t0 V
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0−) = Vs(0−) = 0 V
v(0+) ≃ v(0−) = 0 V
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
As t→∞ , i→ 0 → v(∞) = Vs(∞) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., A = V0 , B = −V0
t = 0+: 0 = A + B ,
t→∞: V0 = B .
v(t) = V0 [1− exp(−t/τ)]
i
t0 V
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
v(0−) = Vs(0−) = V0
v(0+) ≃ v(0−) = V0
As t→∞ , i→ 0 → v(∞) = Vs(∞) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A + B ,
i.e., A = V0 , B = 0
t→∞: 0 = B .
v(t) = V0 exp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
i
t0 V
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0−) = Vs(0−) = 0 V
v(0+) ≃ v(0−) = 0 V
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
As t→∞ , i→ 0 → v(∞) = Vs(∞) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., A = V0 , B = −V0
t = 0+: 0 = A + B ,
t→∞: V0 = B .
v(t) = V0 [1− exp(−t/τ)]
i
t0 V
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
v(0−) = Vs(0−) = V0
v(0+) ≃ v(0−) = V0
As t→∞ , i→ 0 → v(∞) = Vs(∞) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A + B ,
i.e., A = V0 , B = 0
t→∞: 0 = B .
v(t) = V0 exp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
i
t0 V
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0−) = Vs(0−) = 0 V
v(0+) ≃ v(0−) = 0 V
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
As t→∞ , i→ 0 → v(∞) = Vs(∞) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., A = V0 , B = −V0
t = 0+: 0 = A + B ,
t→∞: V0 = B .
v(t) = V0 [1− exp(−t/τ)]
i
t0 V
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(−t/τ) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0−)
because Eq. (A) applies only for t > 0.
v(0−) = Vs(0−) = V0
v(0+) ≃ v(0−) = V0
As t→∞ , i→ 0 → v(∞) = Vs(∞) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A + B ,
i.e., A = V0 , B = 0
t→∞: 0 = B .
v(t) = V0 exp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
i
t0 V
Cv
R
Vs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [1− exp(−t/τ)]
=CV0
τexp(−t/τ) =
V0
Rexp(−t/τ)
Using these conditions, we obtain
(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .
t = 0+: v = 0 , Vs = V0 ⇒ i(0+) = V0/R .
t→∞: i(t) = 0 .
A′ =V0
R, B′ = 0 ⇒ i(t) =
V0
Rexp(−t/τ)
i
t0 V
Cv
R
Vs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [exp(−t/τ)]
= −CV0
τexp(−t/τ) = −V0
Rexp(−t/τ)
Using these conditions, we obtain
(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .
t→∞: i(t) = 0 .
t = 0+: v = V0 , Vs = 0 ⇒ i(0+) = −V0/R .
A′ = −V0
R, B′ = 0 ⇒ i(t) = −V0
Rexp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
i
t0 V
Cv
R
Vs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [1− exp(−t/τ)]
=CV0
τexp(−t/τ) =
V0
Rexp(−t/τ)
Using these conditions, we obtain
(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .
t = 0+: v = 0 , Vs = V0 ⇒ i(0+) = V0/R .
t→∞: i(t) = 0 .
A′ =V0
R, B′ = 0 ⇒ i(t) =
V0
Rexp(−t/τ)
i
t0 V
Cv
R
Vs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [exp(−t/τ)]
= −CV0
τexp(−t/τ) = −V0
Rexp(−t/τ)
Using these conditions, we obtain
(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .
t→∞: i(t) = 0 .
t = 0+: v = V0 , Vs = 0 ⇒ i(0+) = −V0/R .
A′ = −V0
R, B′ = 0 ⇒ i(t) = −V0
Rexp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
i
t0 V
Cv
R
Vs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [1− exp(−t/τ)]
=CV0
τexp(−t/τ) =
V0
Rexp(−t/τ)
Using these conditions, we obtain
(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .
t = 0+: v = 0 , Vs = V0 ⇒ i(0+) = V0/R .
t→∞: i(t) = 0 .
A′ =V0
R, B′ = 0 ⇒ i(t) =
V0
Rexp(−t/τ)
i
t0 V
Cv
R
Vs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [exp(−t/τ)]
= −CV0
τexp(−t/τ) = −V0
Rexp(−t/τ)
Using these conditions, we obtain
(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .
t→∞: i(t) = 0 .
t = 0+: v = V0 , Vs = 0 ⇒ i(0+) = −V0/R .
A′ = −V0
R, B′ = 0 ⇒ i(t) = −V0
Rexp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
i
t0 V
Cv
R
Vs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [1− exp(−t/τ)]
=CV0
τexp(−t/τ) =
V0
Rexp(−t/τ)
Using these conditions, we obtain
(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .
t = 0+: v = 0 , Vs = V0 ⇒ i(0+) = V0/R .
t→∞: i(t) = 0 .
A′ =V0
R, B′ = 0 ⇒ i(t) =
V0
Rexp(−t/τ)
i
t0 V
Cv
R
Vs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [exp(−t/τ)]
= −CV0
τexp(−t/τ) = −V0
Rexp(−t/τ)
Using these conditions, we obtain
(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .
t→∞: i(t) = 0 .
t = 0+: v = V0 , Vs = 0 ⇒ i(0+) = −V0/R .
A′ = −V0
R, B′ = 0 ⇒ i(t) = −V0
Rexp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
i
t0 V
Cv
R
Vs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [1− exp(−t/τ)]
=CV0
τexp(−t/τ) =
V0
Rexp(−t/τ)
Using these conditions, we obtain
(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .
t = 0+: v = 0 , Vs = V0 ⇒ i(0+) = V0/R .
t→∞: i(t) = 0 .
A′ =V0
R, B′ = 0 ⇒ i(t) =
V0
Rexp(−t/τ)
i
t0 V
Cv
R
Vs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [exp(−t/τ)]
= −CV0
τexp(−t/τ) = −V0
Rexp(−t/τ)
Using these conditions, we obtain
(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .
t→∞: i(t) = 0 .
t = 0+: v = V0 , Vs = 0 ⇒ i(0+) = −V0/R .
A′ = −V0
R, B′ = 0 ⇒ i(t) = −V0
Rexp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
i
t0 V
Cv
R
Vs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [1− exp(−t/τ)]
=CV0
τexp(−t/τ) =
V0
Rexp(−t/τ)
Using these conditions, we obtain
(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .
t = 0+: v = 0 , Vs = V0 ⇒ i(0+) = V0/R .
t→∞: i(t) = 0 .
A′ =V0
R, B′ = 0 ⇒ i(t) =
V0
Rexp(−t/τ)
i
t0 V
Cv
R
Vs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [exp(−t/τ)]
= −CV0
τexp(−t/τ) = −V0
Rexp(−t/τ)
Using these conditions, we obtain
(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 .
t→∞: i(t) = 0 .
t = 0+: v = V0 , Vs = 0 ⇒ i(0+) = −V0/R .
A′ = −V0
R, B′ = 0 ⇒ i(t) = −V0
Rexp(−t/τ)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
i
t0 V
C = 1 µFvVs
R = 1 kVs
5 V
v(t) = V0 [1− exp(−t/τ)]
i(t) =V0
Rexp(−t/τ)
0
5
v (V
olts
) v
Vs
5
0
i (m
A)
time (msec)
−2 0 2 4 6 8
i
t0 V
C = 1 µFvVs
R = 1 kVs
5 V
v(t) = V0 exp(−t/τ)
i(t) = −V0
Rexp(−t/τ)
v (V
olts
)
5
0
v
Vs
i (m
A)
0
−5
time (msec)
−2 0 2 4 6 8
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
i
t0 V
C = 1 µFvVs
R = 1 kVs
5 V
v(t) = V0 [1− exp(−t/τ)]
i(t) =V0
Rexp(−t/τ)
0
5
v (V
olts
) v
Vs
5
0
i (m
A)
time (msec)
−2 0 2 4 6 8
i
t0 V
C = 1 µFvVs
R = 1 kVs
5 V
v(t) = V0 exp(−t/τ)
i(t) = −V0
Rexp(−t/τ)
v (V
olts
)
5
0
v
Vs
i (m
A)
0
−5
time (msec)
−2 0 2 4 6 8
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
i
t0 V
C = 1 µFvVs
R = 1 kVs
5 V
v(t) = V0 [1− exp(−t/τ)]
i(t) =V0
Rexp(−t/τ)
0
5
v (V
olts
) v
Vs
5
0
i (m
A)
time (msec)
−2 0 2 4 6 8
i
t0 V
C = 1 µFvVs
R = 1 kVs
5 V
v(t) = V0 exp(−t/τ)
i(t) = −V0
Rexp(−t/τ)
v (V
olts
)
5
0
v
Vs
i (m
A)
0
−5
time (msec)
−2 0 2 4 6 8
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
i
t0 V
C = 1 µFvVs
R = 1 kVs
5 V
v(t) = V0 [1− exp(−t/τ)]
i(t) =V0
Rexp(−t/τ)
0
5
v (V
olts
) v
Vs
5
0
i (m
A)
time (msec)
−2 0 2 4 6 8
i
t0 V
C = 1 µFvVs
R = 1 kVs
5 V
v(t) = V0 exp(−t/τ)
i(t) = −V0
Rexp(−t/τ)
v (V
olts
)
5
0
v
Vs
i (m
A)
0
−5
time (msec)
−2 0 2 4 6 8
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
i
t0 V
C = 1 µFvVs
R = 1 kVs
5 V
v(t) = V0 [1− exp(−t/τ)]
i(t) =V0
Rexp(−t/τ)
0
5
v (V
olts
) v
Vs
5
0
i (m
A)
time (msec)
−2 0 2 4 6 8
i
t0 V
C = 1 µFvVs
R = 1 kVs
5 V
v(t) = V0 exp(−t/τ)
i(t) = −V0
Rexp(−t/τ)
v (V
olts
)
5
0
v
Vs
i (m
A)
0
−5
time (msec)
−2 0 2 4 6 8
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
i
t0 V
C = 1 µFvVs
R = 1 kVs
5 V
v(t) = V0 [1− exp(−t/τ)]
i(t) =V0
Rexp(−t/τ)
0
5
v (V
olts
) v
Vs
5
0
i (m
A)
time (msec)
−2 0 2 4 6 8
i
t0 V
C = 1 µFvVs
R = 1 kVs
5 V
v(t) = V0 exp(−t/τ)
i(t) = −V0
Rexp(−t/τ)
v (V
olts
)
5
0
v
Vs
i (m
A)
0
−5
time (msec)
−2 0 2 4 6 8
M. B. Patil, IIT Bombay
Significance of the time constant (τ)
x e−x 1− e−x
0.0 1.0 0.0
1.0 0.3679 0.6321
2.0 0.1353 0.8647
3.0 4.9787×10−2 0.9502
4.0 1.8315×10−2 0.9817
5.0 6.7379×10−3 0.9933
* For x = 5, e−x ' 0, 1− e−x ' 1.
* In RC circuits, x = t/τ ⇒ When t = 5 τ , the charging (or discharging) processis almost complete.
0
1
x
0 1 2 3 4 5 6
exp(−x)
1− exp(−x)
M. B. Patil, IIT Bombay
Significance of the time constant (τ)
x e−x 1− e−x
0.0 1.0 0.0
1.0 0.3679 0.6321
2.0 0.1353 0.8647
3.0 4.9787×10−2 0.9502
4.0 1.8315×10−2 0.9817
5.0 6.7379×10−3 0.9933
* For x = 5, e−x ' 0, 1− e−x ' 1.
* In RC circuits, x = t/τ ⇒ When t = 5 τ , the charging (or discharging) processis almost complete.
0
1
x
0 1 2 3 4 5 6
exp(−x)
1− exp(−x)
M. B. Patil, IIT Bombay
Significance of the time constant (τ)
x e−x 1− e−x
0.0 1.0 0.0
1.0 0.3679 0.6321
2.0 0.1353 0.8647
3.0 4.9787×10−2 0.9502
4.0 1.8315×10−2 0.9817
5.0 6.7379×10−3 0.9933
* For x = 5, e−x ' 0, 1− e−x ' 1.
* In RC circuits, x = t/τ ⇒ When t = 5 τ , the charging (or discharging) processis almost complete.
0
1
x
0 1 2 3 4 5 6
exp(−x)
1− exp(−x)
M. B. Patil, IIT Bombay
Significance of the time constant (τ)
x e−x 1− e−x
0.0 1.0 0.0
1.0 0.3679 0.6321
2.0 0.1353 0.8647
3.0 4.9787×10−2 0.9502
4.0 1.8315×10−2 0.9817
5.0 6.7379×10−3 0.9933
* For x = 5, e−x ' 0, 1− e−x ' 1.
* In RC circuits, x = t/τ ⇒ When t = 5 τ , the charging (or discharging) processis almost complete.
0
1
x
0 1 2 3 4 5 6
exp(−x)
1− exp(−x)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
i i
time (msec) time (msec)
t0 V t0 V
0
5
v (V
olts
)
v (V
olts
)
5
0
−1 0 1 2 3 4 5 6 −1 0 1 2 3 4 5 6
C = 1 µF C = 1 µFv vVs Vs
R RVs Vs
5 V 5 V
R = 1 kΩ
R = 100 Ω
R = 1 kΩ
R = 100 Ω
v(t) = V0 exp(−t/τ)v(t) = V0 [1− exp(−t/τ)]
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
t1t0
R2
R1
Vs
VsR1 = 10 Ω
R2 = 40 Ω
L = 0.8 H
t0 = 0
t1 = 0.1 s
i(0) = 0 A, Find i(t).
(1) t < t0
(2) t0 < t < t1
(3) t > t1
There are three intervals of constant Vs:
R2
R1
Vs
RTh seen by L is the same in all intervals:
τ = L/RTh
= 0.1 s
= 0.8 H/8Ω
RTh = R1 ‖ R2 = 8 Ω
⇒ i(t−0 ) = 0 A⇒ i(t+0 ) = 0 A .
At t = t−0 , v = 0 V, Vs = 0 V .
10 V
t
v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .
If Vs did not change at t = t1,
we would have
t1t0
Vs
i(t), t > 0 (See next slide).
Using i(t+0 ) and i(∞), we can obtain
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
t1t0
R2
R1
Vs
VsR1 = 10 Ω
R2 = 40 Ω
L = 0.8 H
t0 = 0
t1 = 0.1 s
i(0) = 0 A, Find i(t).
(1) t < t0
(2) t0 < t < t1
(3) t > t1
There are three intervals of constant Vs:
R2
R1
Vs
RTh seen by L is the same in all intervals:
τ = L/RTh
= 0.1 s
= 0.8 H/8Ω
RTh = R1 ‖ R2 = 8 Ω
⇒ i(t−0 ) = 0 A⇒ i(t+0 ) = 0 A .
At t = t−0 , v = 0 V, Vs = 0 V .
10 V
t
v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .
If Vs did not change at t = t1,
we would have
t1t0
Vs
i(t), t > 0 (See next slide).
Using i(t+0 ) and i(∞), we can obtain
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
t1t0
R2
R1
Vs
VsR1 = 10 Ω
R2 = 40 Ω
L = 0.8 H
t0 = 0
t1 = 0.1 s
i(0) = 0 A, Find i(t).
(1) t < t0
(2) t0 < t < t1
(3) t > t1
There are three intervals of constant Vs:
R2
R1
Vs
RTh seen by L is the same in all intervals:
τ = L/RTh
= 0.1 s
= 0.8 H/8Ω
RTh = R1 ‖ R2 = 8 Ω
⇒ i(t−0 ) = 0 A⇒ i(t+0 ) = 0 A .
At t = t−0 , v = 0 V, Vs = 0 V .
10 V
t
v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .
If Vs did not change at t = t1,
we would have
t1t0
Vs
i(t), t > 0 (See next slide).
Using i(t+0 ) and i(∞), we can obtain
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
t1t0
R2
R1
Vs
VsR1 = 10 Ω
R2 = 40 Ω
L = 0.8 H
t0 = 0
t1 = 0.1 s
i(0) = 0 A, Find i(t).
(1) t < t0
(2) t0 < t < t1
(3) t > t1
There are three intervals of constant Vs:
R2
R1
Vs
RTh seen by L is the same in all intervals:
τ = L/RTh
= 0.1 s
= 0.8 H/8Ω
RTh = R1 ‖ R2 = 8 Ω
⇒ i(t−0 ) = 0 A⇒ i(t+0 ) = 0 A .
At t = t−0 , v = 0 V, Vs = 0 V .
10 V
t
v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .
If Vs did not change at t = t1,
we would have
t1t0
Vs
i(t), t > 0 (See next slide).
Using i(t+0 ) and i(∞), we can obtain
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
t1t0
R2
R1
Vs
VsR1 = 10 Ω
R2 = 40 Ω
L = 0.8 H
t0 = 0
t1 = 0.1 s
i(0) = 0 A, Find i(t).
(1) t < t0
(2) t0 < t < t1
(3) t > t1
There are three intervals of constant Vs:
R2
R1
Vs
RTh seen by L is the same in all intervals:
τ = L/RTh
= 0.1 s
= 0.8 H/8Ω
RTh = R1 ‖ R2 = 8 Ω
⇒ i(t−0 ) = 0 A⇒ i(t+0 ) = 0 A .
At t = t−0 , v = 0 V, Vs = 0 V .
10 V
t
v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .
If Vs did not change at t = t1,
we would have
t1t0
Vs
i(t), t > 0 (See next slide).
Using i(t+0 ) and i(∞), we can obtain
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
t1t0
R2
R1
Vs
VsR1 = 10 Ω
R2 = 40 Ω
L = 0.8 H
t0 = 0
t1 = 0.1 s
i(0) = 0 A, Find i(t).
(1) t < t0
(2) t0 < t < t1
(3) t > t1
There are three intervals of constant Vs:
R2
R1
Vs
RTh seen by L is the same in all intervals:
τ = L/RTh
= 0.1 s
= 0.8 H/8Ω
RTh = R1 ‖ R2 = 8 Ω
⇒ i(t−0 ) = 0 A⇒ i(t+0 ) = 0 A .
At t = t−0 , v = 0 V, Vs = 0 V .
10 V
t
v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A .
If Vs did not change at t = t1,
we would have
t1t0
Vs
i(t), t > 0 (See next slide).
Using i(t+0 ) and i(∞), we can obtain
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
i (A
mp)
1
0
0time (sec)
0.2 0.4 0.6 0.8
t1t0
R2
R1
Vs
VsR1 = 10 Ω
R2 = 40 Ω
L = 0.8 H
t0 = 0
t1 = 0.1 s
and we need to work out the
solution for t > t1 separately.
In reality, Vs changes at t = t1,
Consider t > t1.
For t0 < t < t1, i(t) = 1− exp(−t/τ) Amp.
i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).
i(∞) = 0 A.
Let i(t) = A exp(−t/τ) + B.
It is convenient to rewrite i(t) as
i(t) = A′ exp[−(t− t1)/τ ] + B.
Using i(t+1 ) and i(∞), we get
i(t) = 0.693 exp[−(t− t1)/τ ] A.
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
i (A
mp)
1
0
0time (sec)
0.2 0.4 0.6 0.8
t1t0
R2
R1
Vs
VsR1 = 10 Ω
R2 = 40 Ω
L = 0.8 H
t0 = 0
t1 = 0.1 s
and we need to work out the
solution for t > t1 separately.
In reality, Vs changes at t = t1,
Consider t > t1.
For t0 < t < t1, i(t) = 1− exp(−t/τ) Amp.
i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).
i(∞) = 0 A.
Let i(t) = A exp(−t/τ) + B.
It is convenient to rewrite i(t) as
i(t) = A′ exp[−(t− t1)/τ ] + B.
Using i(t+1 ) and i(∞), we get
i(t) = 0.693 exp[−(t− t1)/τ ] A.
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
i (A
mp)
1
0
0time (sec)
0.2 0.4 0.6 0.8
t1t0
R2
R1
Vs
VsR1 = 10 Ω
R2 = 40 Ω
L = 0.8 H
t0 = 0
t1 = 0.1 s
and we need to work out the
solution for t > t1 separately.
In reality, Vs changes at t = t1,
Consider t > t1.
For t0 < t < t1, i(t) = 1− exp(−t/τ) Amp.
i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).
i(∞) = 0 A.
Let i(t) = A exp(−t/τ) + B.
It is convenient to rewrite i(t) as
i(t) = A′ exp[−(t− t1)/τ ] + B.
Using i(t+1 ) and i(∞), we get
i(t) = 0.693 exp[−(t− t1)/τ ] A.
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
i (A
mp)
1
0
0time (sec)
0.2 0.4 0.6 0.8
t1t0
R2
R1
Vs
VsR1 = 10 Ω
R2 = 40 Ω
L = 0.8 H
t0 = 0
t1 = 0.1 s
and we need to work out the
solution for t > t1 separately.
In reality, Vs changes at t = t1,
Consider t > t1.
For t0 < t < t1, i(t) = 1− exp(−t/τ) Amp.
i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).
i(∞) = 0 A.
Let i(t) = A exp(−t/τ) + B.
It is convenient to rewrite i(t) as
i(t) = A′ exp[−(t− t1)/τ ] + B.
Using i(t+1 ) and i(∞), we get
i(t) = 0.693 exp[−(t− t1)/τ ] A.
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
i (A
mp)
1
0
0time (sec)
0.2 0.4 0.6 0.8
t1t0
R2
R1
Vs
VsR1 = 10 Ω
R2 = 40 Ω
L = 0.8 H
t0 = 0
t1 = 0.1 s
and we need to work out the
solution for t > t1 separately.
In reality, Vs changes at t = t1,
Consider t > t1.
For t0 < t < t1, i(t) = 1− exp(−t/τ) Amp.
i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).
i(∞) = 0 A.
Let i(t) = A exp(−t/τ) + B.
It is convenient to rewrite i(t) as
i(t) = A′ exp[−(t− t1)/τ ] + B.
Using i(t+1 ) and i(∞), we get
i(t) = 0.693 exp[−(t− t1)/τ ] A.
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
i (A
mp)
1
0
0time (sec)
0.2 0.4 0.6 0.8
t1t0
R2
R1
Vs
VsR1 = 10 Ω
R2 = 40 Ω
L = 0.8 H
t0 = 0
t1 = 0.1 s
and we need to work out the
solution for t > t1 separately.
In reality, Vs changes at t = t1,
Consider t > t1.
For t0 < t < t1, i(t) = 1− exp(−t/τ) Amp.
i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).
i(∞) = 0 A.
Let i(t) = A exp(−t/τ) + B.
It is convenient to rewrite i(t) as
i(t) = A′ exp[−(t− t1)/τ ] + B.
Using i(t+1 ) and i(∞), we get
i(t) = 0.693 exp[−(t− t1)/τ ] A.
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
i (A
mp)
1
0
0time (sec)
0.2 0.4 0.6 0.8
t1t0
R2
R1
Vs
VsR1 = 10 Ω
R2 = 40 Ω
L = 0.8 H
t0 = 0
t1 = 0.1 s
and we need to work out the
solution for t > t1 separately.
In reality, Vs changes at t = t1,
Consider t > t1.
For t0 < t < t1, i(t) = 1− exp(−t/τ) Amp.
i(t+1 ) = i(t−1 ) = 1− e−1 = 0.632 A (Note: t1/τ = 1).
i(∞) = 0 A.
Let i(t) = A exp(−t/τ) + B.
It is convenient to rewrite i(t) as
i(t) = A′ exp[−(t− t1)/τ ] + B.
Using i(t+1 ) and i(∞), we get
i(t) = 0.693 exp[−(t− t1)/τ ] A.
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
i (A
mp)
1
0
0 0.2 0.4 0.6 0.8time (sec)
i(t) = 0.693 exp[−(t− t1)/τ ] A.
t1t0
R2
R1
Vs
VsR1 = 10 Ω
R2 = 40 Ω
L = 0.8 H
t0 = 0
t1 = 0.1 s
(SEQUEL file: ee101_rl1.sqproj)
0
1
0 0.2 0.4 0.6 0.8
i (A
mp)
time (sec)
Combining the solutions for t0 < t < t1 and t > t1,
we get
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
i (A
mp)
1
0
0 0.2 0.4 0.6 0.8time (sec)
i(t) = 0.693 exp[−(t− t1)/τ ] A.
t1t0
R2
R1
Vs
VsR1 = 10 Ω
R2 = 40 Ω
L = 0.8 H
t0 = 0
t1 = 0.1 s
(SEQUEL file: ee101_rl1.sqproj)
0
1
0 0.2 0.4 0.6 0.8
i (A
mp)
time (sec)
Combining the solutions for t0 < t < t1 and t > t1,
we get
M. B. Patil, IIT Bombay
RC circuit: example
1 k
t=0i
6 V
5 kR1 5 µF
ic
vc
R3 = 5 k
R2
AND
i i
5 k 1 k
6 V
5 k 5 k
5 k
5 µF
ic
t < 0
5 µF
ic
vc
t > 0
vc
t = 0−: capacitor is an open circuit, ⇒ i(0−) = 6 V/(5 k + 1 k) = 1 mA.
vc(0−) = 6 V− 1 mA× R2 = 5 V⇒ vc(0
+) = 5 V.
⇒ i(0+) = 5 V/(5 k + 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5 µF = 50 ms.
i(t) = 0.5 exp(-t/τ) mA.
Using i(0+) and i(∞) = 0 A, we get
0
1
i (mA)
time (sec) 0 0.5
0
−0.50
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: example
1 k
t=0i
6 V
5 kR1 5 µF
ic
vc
R3 = 5 k
R2 AND
i i
5 k 1 k
6 V
5 k 5 k
5 k
5 µF
ic
t < 0
5 µF
ic
vc
t > 0
vc
t = 0−: capacitor is an open circuit, ⇒ i(0−) = 6 V/(5 k + 1 k) = 1 mA.
vc(0−) = 6 V− 1 mA× R2 = 5 V⇒ vc(0
+) = 5 V.
⇒ i(0+) = 5 V/(5 k + 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5 µF = 50 ms.
i(t) = 0.5 exp(-t/τ) mA.
Using i(0+) and i(∞) = 0 A, we get
0
1
i (mA)
time (sec) 0 0.5
0
−0.50
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: example
1 k
t=0i
6 V
5 kR1 5 µF
ic
vc
R3 = 5 k
R2 AND
i i
5 k 1 k
6 V
5 k 5 k
5 k
5 µF
ic
t < 0
5 µF
ic
vc
t > 0
vc
t = 0−: capacitor is an open circuit, ⇒ i(0−) = 6 V/(5 k + 1 k) = 1 mA.
vc(0−) = 6 V− 1 mA× R2 = 5 V⇒ vc(0
+) = 5 V.
⇒ i(0+) = 5 V/(5 k + 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5 µF = 50 ms.
i(t) = 0.5 exp(-t/τ) mA.
Using i(0+) and i(∞) = 0 A, we get
0
1
i (mA)
time (sec) 0 0.5
0
−0.50
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: example
1 k
t=0i
6 V
5 kR1 5 µF
ic
vc
R3 = 5 k
R2 AND
i i
5 k 1 k
6 V
5 k 5 k
5 k
5 µF
ic
t < 0
5 µF
ic
vc
t > 0
vc
t = 0−: capacitor is an open circuit, ⇒ i(0−) = 6 V/(5 k + 1 k) = 1 mA.
vc(0−) = 6 V− 1 mA× R2 = 5 V⇒ vc(0
+) = 5 V.
⇒ i(0+) = 5 V/(5 k + 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5 µF = 50 ms.
i(t) = 0.5 exp(-t/τ) mA.
Using i(0+) and i(∞) = 0 A, we get
0
1
i (mA)
time (sec) 0 0.5
0
−0.50
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: example
1 k
t=0i
6 V
5 kR1 5 µF
ic
vc
R3 = 5 k
R2 AND
i i
5 k 1 k
6 V
5 k 5 k
5 k
5 µF
ic
t < 0
5 µF
ic
vc
t > 0
vc
t = 0−: capacitor is an open circuit, ⇒ i(0−) = 6 V/(5 k + 1 k) = 1 mA.
vc(0−) = 6 V− 1 mA× R2 = 5 V⇒ vc(0
+) = 5 V.
⇒ i(0+) = 5 V/(5 k + 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5 µF = 50 ms.
i(t) = 0.5 exp(-t/τ) mA.
Using i(0+) and i(∞) = 0 A, we get
0
1
i (mA)
time (sec) 0 0.5
0
−0.50
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: example
1 k
t=0i
6 V
5 kR1 5 µF
ic
vc
R3 = 5 k
R2 AND
i i
5 k 1 k
6 V
5 k 5 k
5 k
5 µF
ic
t < 0
5 µF
ic
vc
t > 0
vc
t = 0−: capacitor is an open circuit, ⇒ i(0−) = 6 V/(5 k + 1 k) = 1 mA.
vc(0−) = 6 V− 1 mA× R2 = 5 V⇒ vc(0
+) = 5 V.
⇒ i(0+) = 5 V/(5 k + 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5 µF = 50 ms.
i(t) = 0.5 exp(-t/τ) mA.
Using i(0+) and i(∞) = 0 A, we get
0
1
i (mA)
time (sec) 0 0.5
0
−0.50
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: example
1 k
t=0i
6 V
5 kR1 5 µF
ic
vc
R3 = 5 k
R2 AND
i i
5 k 1 k
6 V
5 k 5 k
5 k
5 µF
ic
t < 0
5 µF
ic
vc
t > 0
vc
t = 0−: capacitor is an open circuit, ⇒ i(0−) = 6 V/(5 k + 1 k) = 1 mA.
vc(0−) = 6 V− 1 mA× R2 = 5 V⇒ vc(0
+) = 5 V.
⇒ i(0+) = 5 V/(5 k + 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5 µF = 50 ms.
i(t) = 0.5 exp(-t/τ) mA.
Using i(0+) and i(∞) = 0 A, we get
0
1
i (mA)
time (sec) 0 0.5
0
−0.50
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: example
1 k
t=0i
6 V
5 kR1 5 µF
ic
vc
R3 = 5 k
R2 AND
i i
5 k 1 k
6 V
5 k 5 k
5 k
5 µF
ic
t < 0
5 µF
ic
vc
t > 0
vc
t = 0−: capacitor is an open circuit, ⇒ i(0−) = 6 V/(5 k + 1 k) = 1 mA.
vc(0−) = 6 V− 1 mA× R2 = 5 V⇒ vc(0
+) = 5 V.
⇒ i(0+) = 5 V/(5 k + 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5 µF = 50 ms.
i(t) = 0.5 exp(-t/τ) mA.
Using i(0+) and i(∞) = 0 A, we get
0
1
i (mA)
time (sec) 0 0.5
0
−0.50
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: example
1 k
t=0i
6 V
5 kR1 5 µF
ic
vc
R3 = 5 k
R2 AND
i i
5 k 1 k
6 V
5 k 5 k
5 k
5 µF
ic
t < 0
5 µF
ic
vc
t > 0
vc
t = 0−: capacitor is an open circuit, ⇒ i(0−) = 6 V/(5 k + 1 k) = 1 mA.
vc(0−) = 6 V− 1 mA× R2 = 5 V⇒ vc(0
+) = 5 V.
⇒ i(0+) = 5 V/(5 k + 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5 µF = 50 ms.
i(t) = 0.5 exp(-t/τ) mA.
Using i(0+) and i(∞) = 0 A, we get
0
1
i (mA)
time (sec) 0 0.5
0
−0.50
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: example
1 k
t=0i
6 V
5 kR1 5 µF
ic
vc
R3 = 5 k
R2 AND
i i
5 k 1 k
6 V
5 k 5 k
5 k
5 µF
ic
t < 0
5 µF
ic
vc
t > 0
vc
t = 0−: capacitor is an open circuit, ⇒ i(0−) = 6 V/(5 k + 1 k) = 1 mA.
vc(0−) = 6 V− 1 mA× R2 = 5 V⇒ vc(0
+) = 5 V.
⇒ i(0+) = 5 V/(5 k + 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/τ) + B for t > 0, with τ = 10 k× 5 µF = 50 ms.
i(t) = 0.5 exp(-t/τ) mA.
Using i(0+) and i(∞) = 0 A, we get
0
1
i (mA)
time (sec) 0 0.5
0
−0.50
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuits: home work
i1i2
vc
ic
10 Ω
10 Ω
200 µF10V
* Given vc (0) = 0 V , find vc (t) for t > 0. Using this vc (t), find i1, i2, ic for t > 0.Plot vc , i1, i2, ic versus t.
* Find i1, i2, ic directly (i.e., without getting vc ) by finding the initial and finalconditions for each of them (i1(0+) and i1(∞), etc.) and then using them tocompute the coefficients in the general expression,x(t) = A exp(−t/τ) + B.
* Verify your results with SEQUEL (file: ee101 rc3.sqproj).
M. B. Patil, IIT Bombay
RC circuits: home work
i1i2
vc
ic
10 Ω
10 Ω
200 µF10V
* Given vc (0) = 0 V , find vc (t) for t > 0. Using this vc (t), find i1, i2, ic for t > 0.Plot vc , i1, i2, ic versus t.
* Find i1, i2, ic directly (i.e., without getting vc ) by finding the initial and finalconditions for each of them (i1(0+) and i1(∞), etc.) and then using them tocompute the coefficients in the general expression,x(t) = A exp(−t/τ) + B.
* Verify your results with SEQUEL (file: ee101 rc3.sqproj).
M. B. Patil, IIT Bombay
RC circuits: home work
i1i2
vc
ic
10 Ω
10 Ω
200 µF10V
* Given vc (0) = 0 V , find vc (t) for t > 0. Using this vc (t), find i1, i2, ic for t > 0.Plot vc , i1, i2, ic versus t.
* Find i1, i2, ic directly (i.e., without getting vc ) by finding the initial and finalconditions for each of them (i1(0+) and i1(∞), etc.) and then using them tocompute the coefficients in the general expression,x(t) = A exp(−t/τ) + B.
* Verify your results with SEQUEL (file: ee101 rc3.sqproj).
M. B. Patil, IIT Bombay
RC circuits: home work
i1i2
vc
ic
10 Ω
10 Ω
200 µF10V
* Given vc (0) = 0 V , find vc (t) for t > 0. Using this vc (t), find i1, i2, ic for t > 0.Plot vc , i1, i2, ic versus t.
* Find i1, i2, ic directly (i.e., without getting vc ) by finding the initial and finalconditions for each of them (i1(0+) and i1(∞), etc.) and then using them tocompute the coefficients in the general expression,x(t) = A exp(−t/τ) + B.
* Verify your results with SEQUEL (file: ee101 rc3.sqproj).
M. B. Patil, IIT Bombay
RC circuits: home work
t=0 i1
vx 1 mF5Ω vc
ic
2 Ω 3 Ω
0.1 vx24V
* Find vc (0−), vc (∞).
* Find RTh as seen by the capacitor for t > 0.
* Solve for vc (t) and i1(t), t > 0.
* Verify your results with SEQUEL (file: ee101 rc4.sqproj).
M. B. Patil, IIT Bombay
RC circuits: home work
t=0 i1
vx 1 mF5Ω vc
ic
2 Ω 3 Ω
0.1 vx24V
* Find vc (0−), vc (∞).
* Find RTh as seen by the capacitor for t > 0.
* Solve for vc (t) and i1(t), t > 0.
* Verify your results with SEQUEL (file: ee101 rc4.sqproj).
M. B. Patil, IIT Bombay
RC circuits: home work
t=0 i1
vx 1 mF5Ω vc
ic
2 Ω 3 Ω
0.1 vx24V
* Find vc (0−), vc (∞).
* Find RTh as seen by the capacitor for t > 0.
* Solve for vc (t) and i1(t), t > 0.
* Verify your results with SEQUEL (file: ee101 rc4.sqproj).
M. B. Patil, IIT Bombay
RC circuits: home work
t=0 i1
vx 1 mF5Ω vc
ic
2 Ω 3 Ω
0.1 vx24V
* Find vc (0−), vc (∞).
* Find RTh as seen by the capacitor for t > 0.
* Solve for vc (t) and i1(t), t > 0.
* Verify your results with SEQUEL (file: ee101 rc4.sqproj).
M. B. Patil, IIT Bombay
RC circuits: home work
t=0 i1
vx 1 mF5Ω vc
ic
2 Ω 3 Ω
0.1 vx24V
* Find vc (0−), vc (∞).
* Find RTh as seen by the capacitor for t > 0.
* Solve for vc (t) and i1(t), t > 0.
* Verify your results with SEQUEL (file: ee101 rc4.sqproj).
M. B. Patil, IIT Bombay
RL circuits: home work
t=0 i
10 VL=0.1 H5 V
20Ω20Ω
* Find i(0−), i(∞).
* Find RTh as seen by the inductor for t > 0.
* Solve for i(t), t > 0.
* Verify your results with SEQUEL (file: ee101 rl2.sqproj).
M. B. Patil, IIT Bombay
RL circuits: home work
t=0 i
10 VL=0.1 H5 V
20Ω20Ω
* Find i(0−), i(∞).
* Find RTh as seen by the inductor for t > 0.
* Solve for i(t), t > 0.
* Verify your results with SEQUEL (file: ee101 rl2.sqproj).
M. B. Patil, IIT Bombay
RL circuits: home work
t=0 i
10 VL=0.1 H5 V
20Ω20Ω
* Find i(0−), i(∞).
* Find RTh as seen by the inductor for t > 0.
* Solve for i(t), t > 0.
* Verify your results with SEQUEL (file: ee101 rl2.sqproj).
M. B. Patil, IIT Bombay
RL circuits: home work
t=0 i
10 VL=0.1 H5 V
20Ω20Ω
* Find i(0−), i(∞).
* Find RTh as seen by the inductor for t > 0.
* Solve for i(t), t > 0.
* Verify your results with SEQUEL (file: ee101 rl2.sqproj).
M. B. Patil, IIT Bombay
RL circuits: home work
t=0 i
10 VL=0.1 H5 V
20Ω20Ω
* Find i(0−), i(∞).
* Find RTh as seen by the inductor for t > 0.
* Solve for i(t), t > 0.
* Verify your results with SEQUEL (file: ee101 rl2.sqproj).
M. B. Patil, IIT Bombay