Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} 1 PROBLEMS Q#5.1: In the network of the figure, the switch K is closed at t = 0 with the capacitor uncharged. Find values for i, di/dt and d 2 i/dt 2 at t = 0+, for element values as follows: V 100 V R 1000 C 1 F + R V C - Switch is closed at t = 0 (reference time) We know Voltage across capacitor before switching = V C (0-) = 0 V According to the statement under Q#5.1. V C (0+) = V C (0-) = 0 V V 100 i C (0+) = i(0+) = = = 0.1 Amp. R 1000 Element and initial condition Equivalent circuit at t = 0+ C Sc Switch Drop Rise i(0+) Short circuit Drop
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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} 1
PROBLEMSQ#5.1: In the network of the figure, the switch K is closed at t = 0 with the capacitor uncharged. Find values for i, di/dt and d2i/dt2 at t = 0+, for element values as follows:
V 100 VR 1000 C 1 F
+ R
V C-
Switch is closed at t = 0 (reference time)We knowVoltage across capacitor before switching = VC(0-) = 0 VAccording to the statement under Q#5.1.VC(0+) = VC(0-) = 0 V V 100iC(0+) = i(0+) = = = 0.1 Amp. R 1000
Element and initial condition Equivalent circuit at t = 0+
CSc
Switch
Drop
Rise i(0+) Short circuit
Drop
Applying KVL for t 0Sum of voltage rise = sum of voltage drop
1V = iR + idt
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CDifferentiating with respect to ‘t’ di iR + = 0 dt Cdi(0+) -i(0+) = [eq. 1] dt CRBy putting the values of i(0+), C & Rdi(0+) -(0.1) = dt (1 F)(1 k)
di(0+) = -100 Amp/sec dt
Differentiating eq. 1 with respect to ‘t’d2i(0+) -di(0+) 1 = dt2 dt CR
Putting the corresponding values
d2i(0+) = 100, 000 amp/sec2 dt2
Q#5.2: In the given network, K is closed at t = 0 with zero current in the inductor. Find the values of i, di/dt, and d2i/dt2 at t = 0+ if
R 10 L 1 HV 100 V
K+ R
V- L
Key closed at t = 0iL(0+) = iL(0-) = i(0+) = 0 Amp
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According to the statement under Q#5.2:
Drop
Rise Open circuit Drop
i(0+)
Element and initial condition Equivalent circuit at t = 0+ oc
Applying KVL for t 0Sum of voltage rise = sum of voltage drop LdiV = iR + dt Ldi = V – iR dt di V - iR = [eq. 1] dt Ldi(0+) V – i(0+)R = dt LPutting corresponding valuesdi(0+) V – (0)R = dt Ldi(0+) V = dt Ldi(0+) 100 = dt 1di(0+) = 100 Amp/sec dt Differentiating [eq. 1] d2i d V iR = - dt2 dt L L
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d2V2(0+)
= -8 Volt/sec2
dt2
Differentiating eq. 23 d2V2 1 d3V2
+ - 0.1e-t = 020 dt2 20 dt3
At t = 0+3 d2V2(0+) 1 d3V2(0+)
+ - 0.1e-t(0+) = 020 dt2 20 dt3
Putting corresponding values and simplifying
d3V2(0+)
= 26 Volts/sec3 dt3
Q#5.6: The network shown in the accompanying figure is in the steady state with the switch k closed. At t = 0, the switch is opened. Determine the voltage across the switch, VK, and dVK/dt at t = 0+.SOLUTION:
VK
R 1 L 1 HC ½ FV 2 V
Equivalent network before switching
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i(0+) sc
V 2iL(0+) = iL(0-) = i(0+) = = = 2 Amp R 1 1Also VK = VC = idt CdVK i = dt CAt t = 0+dVK i(0+) = dt CdVK 2 = dt (1/2)
dVK
= 4 Volts/sec dt
Q#5.7: In the given network, the switch K is opened at t = 0. At t = 0+, solve for the values of v, dv/dt, and d2v/dt2 if
I 10 AR 1000 C 1 F
V
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Switch is opened at t = 0Equivalent circuit at t = 0-
No current flows through R soVC(0-) = VC(0+) = i(0-)R = V(0+)Herei(0-) = 0VC(0-) = VC(0+) = (0)RVC(0-) = VC(0+) = 0 VoltsFor t 0; according to KCL at VV dV + C = I … (1) R dtAt t = 0+V(0+) dV(0+) + C = I R dtSimplifying
dV(0+) = 107 Volts/sec dt
Differentiating (1) with respect to ‘t’
1 dV d2V+ C = 0
R dt dt2
At t = 0+1 dV(0+) d2V(0+)
+ C = 0R dt dt2
Simplifying
d2V(0+)
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= -1010 Volts/sec2 dt2
Q#5.8: The network shown in the figure has the switch K opened at t = 0. Solve for V, dV/dt, and d2V/dt2 at t = 0+ if
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SoV(0+) = (I)(R) V(0+) = (1)(100)
V(0+) = 100 Volts
For t 0Applying KCL at node VV 1 + Vdt = I … (1)R LDifferentiating (1) with respect to ‘t’1 dV V
+ = 0 … (2) R dt LAt t = 0+1 dV(0+) V(0+)
+ = 0R dt LSimplifying
dV(0+)= -10, 000 Volts/sec
dt
Differentiating (2) with respect to ‘t’1 d2V 1 dV
+ = 0 R dt2 L dtAt t = 0+
1 d2V(0+) 1 dV(0+) + = 0
R dt2 L dtSimplifying
d2V(0+)= 1, 000, 000 Volts/sec2
dt2
Q#5.9: In the network shown in the figure, a steady state is reached with the switch K open. At t = 0, the switch is closed. For the element values given, determine the value of Va(0-) and Va(0+).Circuit diagram:
Vb
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Va
Equivalent circuit before switching:
sc
Req = (10 + 20) 10 (10 + 20)(10)Req =
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(10 + 20) + 10 300Req = 40
Req = 7.5
After simplification Req
i(0-)5 V
Vi(0-) = Req
5i(0-) = 7.5i(0-) = 0.667 Amp.i(0-) = iL(0-)
Va(0-)
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Applying KCL at node Vb
Vb - Va Vb – 5 2+ + = 0
20 10 39Vb – 3Va + 10 = 0 … (ii) Substituting value of Vb from (i) into (ii)9[5Va – 10] – 3Va + 10 = 045Va – 90 – 3Va + 10 = 042Va – 80 = 0
Va(0+) = 1.905 Volts
Q#5.10: In the accompanying figure is shown a network in which a steady state is reached with switch K open. At t = 0, the switch is closed. For the element values given, determine the values of Va(0-) and Va(0+).
CIRCUIT DIAGRAM:Vb
Va
K
Equivalent circuit before switching
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i(0-)5 V
Vi(0-) = Req
5i(0-) = 7.5i(0-) = 0.667 Amp.i(0-) = iL(0-)
iL(0-) = iL(0+) = 0.667 Amp.
Equivalent network at t =
Va
Applying KCL at node Va iL()
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Va – 5 Va Va
+ + = 0 10 20 10After simplification we get
Va() = 2 Volts
V Va
iL() = + 10 20After simplification we get
iL() = 0.6 Amp.
Q#5.12: In the network given in figure p5-10, determine Vb(0+) and Vb() for the conditions stated in Prob. 5-10.At t = 0-, equivalent network is
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VC(0-) = VC(0+) = 5 VoltsAlso equivalent network at t = is
Vb
Va
According to KCL at Va
Va – 5 Va - Vb Va
+ + = 0 10 20 10After simplification we getVa = 0.2Vb + 2 … (i)According to KCL at Vb
Vb – 5 Vb – Va
+ = 0 10 20 After simplification we get3Vb – Va – 10 = 0Putting the value of Va we get
Vb() = 4.286 Volts
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Q#5.13: In the accompanying network, the switch K is closed at t = 0 with zero capacitor voltage and zero inductor current. Solve for (a) V1 and V2 at t = 0+, (b) V1 and V2 at t = , (c) dV1/dt and dV2/dt at t = 0+, (d) d2V2/dt2 at t = 0+.CIRCUIT DIAGRAM
R1
V1
L
C
V2
R2
Equivalent network after switching
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According to statement under Q#5.13
At t = 0-VC(0-) 0 VVC(0+) 0 ViL(0-) 0 AiL(0+) 0 A
Equivalent network at t = 0+V2(0+) = iL(0+)(R2)V2(0+) = (0)(R2)
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V1() = 0 Volts Self justified
(c)
V1
V2
According to KCL at node ‘V1’ V – V1 dV1 1 + C + V1 – V2)dt = 0 … (i) R1 dt LDifferentiating with respect to ‘t’
1 dV dV1 d2V1 1- + C + (V1 – V2) = 0 … (iii)
R1 dt dt dt2 L
According to KCL at node ‘V2’V2 1 + V2 – V1)dt = 0 … (ii) R2 L Differentiating with respect to ‘t’
1 dV2 1
+ (V2 – V1) = 0 … (iV) R2 dt L
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As
V1 = V1 + V2
By putting the value of V1 in (iii) & (iv)
1 dV d(V1 + V2) d2(V1 + V2) 1- + C + (V1 + V2 – V2) = 0
R1 dt dt dt2 L
1 dV dV1 dV2 d2V1 d2V2 1- - + C + + (V1) = 0 … (V)
R1 dt dt dt dt2 dt2 L
1 dV2 1
+ (V2 – V1) = 0 … (iV) R2 dt L
1 dV2 1
+ (V2 – (V1 + V2)) = 0 … (iV) R2 dt L
1 dV2 1
+ (V2 – V1 - V2) = 0 R2 dt L
1 dV2 1
+ (–V1) = 0 … (Vi) R2 dt L
From (V) & (Vi) we can find the values of dV1/dt & dV2/dt.(d)Refer part C.Q#5.14: The network of Prob. 5-13 reaches a steady state with the switch K closed. At a new reference time, t = 0, the switch K is opened. Solve for the quantities specified in the four parts of Prob. 5-13.(a)Equivalent circuit before switching
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At t = 0- ViL(0-) = R1 + R2
VC(0-) = V – iR1(0-)R1
HereiR1(0-) = iL(0-)VC(0-) = V – iL(0-)R1
VR1
VC(0-) = V - R1 + R2
VR2
VC(0-) = R1 + R2
V2 = iR2R2
ViR2(0-) = iR1(0-) =
R1 + R2
VR2
V2 = R1 + R2
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VR2 VR2
V1(0+) = - R1 + R2 R1 + R2
V1(0+) = 0 Volts
(b)Equivalent network at t = At t = capacitor will be fully discharged and acts as an open circuit.
HenceVC() = 0 ViL() = 0 AV2() = iL()R2
V2() = (0)R2
V2() = 0 Volt
(c)For t 0, the equivalent network is
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V1
V2
V2 V1
Applying KCL at node ‘V1’ 1 dV1
(V1 – V2)dt + C = 0 … (i) L dt
d -V2
C (V1 + V2) = … (ii) dt R2
AsV1 = V1 + V2
1 d(V1 + V2) (V1 + V2 – V2)dt + C = 0 L dt 1 d(V1 + V2) V1dt + C = 0 L dtDifferentiating (i) with respect to ‘t’
1 d2V1 d2V2
V1 + C + C = 0 … (iii) L dt2 dt2
Differentiating (ii) with respect to ‘t’
d2 -1 dV2
C (V1 + V2) = … (iV) dt2 R2 dt
After simplification we get the values of dV1/dt & dV2/dt.(d)Refer part c.
Q#5.15: The switch K in the network of the figure is closed at t = 0 connecting the battery to an unenergized network. (a) Determine i, di/dt, and d2i/dt2 at t = 0+. (b) Determine V1, dV1/dt, and d2V1/dt2 at t = 0+.
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L + i(0+)R2 = V0 … (i) dtBy putting corresponding values we get
di(0+) V0
= dt L
Differentiating (i) with respect to ‘t’ d2i di L + R2 = 0 dt2 dtAt t = 0+ d2i(0+) di(0+) L + R2 = 0 dt2 dtSimplifying we get
d2i(0+) -RV0
= dt2 L2
Referring to the network at t = 0+V1(0+) = VR1(0+) = iR1(0+)(R1) V0
iR1(0+) = R1
V0R1
V1(0+) = R1
V1(0+) = V0
(b)AlsoV1 = V0 for all t 0
dV1(0+) d2V1(0+) = = 0 dt dt2
Q#5.16: The network of Prob. 5.15 reaches a steady state under the conditions specified in that problem. At a new reference time, t = 0, the switch K is opened. Solve for the quantities specified in Prob. 5.15 at t = 0+.
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Equivalent circuit before switching
V0
R2
V0
iL(0-) = R2
VC(0-) = VR2(0-) = iL(0-)(R2) V0
VC(0-) = VR2(0-) = (R2) R2
VC(0-) = VR2(0-) = V0 As
VC(0-) = VC(0+) = V0
Equivalent network at t = 0+
+ -
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i(0+)
V0
i(0+) = iL(0-) = iL(0+) = R2
V1(0+) = V0 – VR1(0+)V1(0+) = V0 – iL(0+)R1
V0
V1(0+) = V0 – R1
R2
R2 – R1
V1(0+) = V0 – R2
Equivalent circuit for t 0
R1
Li
C
R2
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Applying KVL around the loop 1 dii(R1 + R2) + idt + L = 0 … (i) C dtAlsoVR1 + VR2 + VL + VC = 0 i(0+)R1 + i(0+)R2 + VL(0+) + VC(0+) = 0
diVL = L dtdi(0+) VL(0+) = dt LPutting corresponding value
di(0+) -V0R1
= dt R2L
Differentiating eq. (i) with respect to ‘t’ 1 dii(R1 + R2) + idt + L = 0 … (i) C dt
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di i d2i(R1 + R2) + + L = 0 dt C dt2
At t = 0+ di(0+) i(0+) d2i(0+)(R1 + R2) + + L = 0 dt C dt2
Here
di(0+) -V0R1
= dt R2L
V0
i(0+) = R2
Putting corresponding values and simplifying
d2i(0+) V0 R1(R1 + R2) 1 = - dt2 R2 L CAlso diV1 = L + iR2
dtDifferentiating with respect to ‘t’
dV1 d2i di = L + R2 dt dt2 dtAt t = 0+dV1(0+) d2i(0+) di(0+) = L + R2 dt dt2 dtBy putting corresponding values and simplifying
dV1(0+) V0 R12 1
= - dt R2 L C
We know -1V1 = idt - iR1
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CDifferentiating with respect to ‘t’
dV1 -i di = - R1 dt C dtDifferentiating with respect to ‘t’d2V1 -di 1 d2i = - R1
dt2 dt C dt2
At t = 0+d2V1(0+) -di(0+) 1 d2i(0+)
= - R1
dt2 dt C dt2
Putting corresponding values and simplifying
d2V1(0+) V0R1 2 R1(R1 + R2) = - dt2 R2L C L
Q#5.17: In the network shown in the accompanying figure, the switch K is changed from a to b at t = 0. Show that at t = 0+, Vi1 = i2 = - R1 + R2 + R3
i3 = 0
a C3
bV R2
+ R3
i1 i2
- i3
R1 L1
C1 C2
Equivalent circuit before switching
L2
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At t = 0-, capacitor C3 is fully charged to voltage V that is VC3(0-) = V and behaves as an open circuit, so current in L1, L2 becomes and other two capacitors also fully charged.iL1(0-) = iL1(0+) = 0 AiL2(0-) = iL2(0+) = 0 AVC1(0-) = VC1(0+) = 0 VVC2(0-) = VC2(0+) = 0 VEquivalent circuit after switching
After simplification we get
+ -
+ -
+ -
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i2
i1
Hence -Vi1 = i2 = R1 + R2 + R3
C1 behaves short circuit being uncharged at t = 0- & L1 behaving open circuit sinceiL(0-) = iL(0+) = 0 Aand i3 = 0 [L2 behaving open circuit].
Q#5.18: In the given network, the capacitor C1 is charged to voltage V0 and the switch K is closed at t = 0. When
R1 2 MV0 1000 VR2 1 MC1 10 FC2 20 F
solve for d2i2/dt2 at t = 0+.
+C1 i1
V0 C2
- i2
R2
R1
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Equivalent circuit before switching
i2
VC1(0-) = V0
VC2(0-) = 0 VEquivalent circuit after switching
K
+
V0 i1 i2
-
VC1(0+) = V0
VC2(0+) = 0 VFor t 0 For loop 1: 1R2(i1 – i2) + i1dt = 0 … (i) C1
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For loop 2: 1R2(i2 – i1) + i2dt + R1i2 = 0 … (ii) C2
1R2(i1 – i2) = - i1dt C1
1R2(i2 – i1) = i1dt … (iii) C1
Taking loop around outsidei2R1 = V0
V0
i2 = … (a) R1
In loop 1 According to KVL:R2(i1 – i2) = V0
R2i1 – R2i2 = V0
Putting the value of i2 and simplifying
V0(R1 + R2)i1 = … (b) R1R2
Putting corresponding values we geti1(0+) 1.5(10-3) Amp.i2(0+) 5(10-4) Amp.
Substituting value of R2(i2 – i1) in eq. (ii)1 1 i1dt + i2dt + R1i2 = 0 C1 C2
Differentiating with respect to ‘t’ i1 di2 i2
+ R1 + = 0 … (iv)C1 dt C2
At t = 0+ i1(0+) di2(0+) i2(0+) + R1 + = 0 C1 dt C2
By putting corresponding values we get
di2(0+) = -8.75(10-5) Amp/sec. dt
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Differentiating eq. (iii) with respect to ‘t’ di2 di1 i1
R2 - R2 =
dt dt C1
At t = 0+ di2(0+) di1(0+) i1(0+)R2 - R2 =
dt dt C1
By putting corresponding values and simplifying
di1(0+) = -2.375(10-4) Amp/sec. dt
Differentiating eq. (iv) with respect to ‘t’1 di1 d2i2 1 di2
Q#5.19: In the circuit shown in the figure, the switch K is closed at t = 0 connecting a voltage, V0sin t, to the parallel RL-RC circuit. Find (a) di1/dt and (b) di2/dt at t = 0+.
i1 i2 + -
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Equivalent circuit after switching
At t 0 Applying KVL around outside loop di2
Ri2 + L = V0sin t … (i) dtApplying KVL around inside loop 1Ri1 + i1dt = V0sin t … (ii) CEquivalent circuit at t = 0+
iL(0+) = iL(0-) = 0 AVC(0+) = 0 V V0sin ti1 = RAt t = 0+
+ -
+ -
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V0sin (0+)i1(0+) = R V0sin 0i1(0+) = Ri1(0+) = 0 AFrom (i)At t = 0+ di2(0+)Ri2(0+) + L = V0sin (0+) … (i) dtBy putting corresponding values we get
di2(0+) = 0 Amp/sec dt
Differentiating eq. (ii) with respect to ‘t’ 1Ri1 + i1dt = V0sin t … (ii) C
di1 i1
R + = V0cos t dt CAt t = 0+ di1(0+) i1(0+)R + = V0cos (0+) dt CBy putting corresponding values & simplifying we get
di1(0+) V0 =
dt R
Q#5.20: In the network shown, a steady state is reached with the switch K open with V 100 VR1 10 R2 20 R3 20 L 1 HC 1 F
. At time t = 0, the switch is closed. (a) Write the integrodifferential equations for the network after the switch is
closed.
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(b) What is the voltage V0 across C before the switch is closed? What is its polarity?
(c) Solve for the initial value of i1 and i2(t = 0+). (d) Solve for the values of di1/dt and di2/dt at t = 0+.(e) What is the value of di1/dt at ?
Circuit diagram:i
i2 i1
Equivalent circuit before switching:
R2
R1
R3
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Simplifying
i
VC(0-)
VC(0-) = iR2(R2) = VR2
Here ViR2 = R1 + R2
VR2
VR2 = R1 + R2
By putting corresponding values we get VC(0-) = VR2 = 66.667 V
ViL(0-) = R1 + R2
iL(0-) = 3.334 A
For t 0
i2 i1
Applying KVL around outside loop di1
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R2i1 + L = V … (i) dtApplying KVL around inside loop 1R3i2 + i2dt = V … (ii) CSince
(V1 – V2) 1 dV1 di(t) + = L R1 dt dtAt t = 0+(V1(0+) – V2(0+)) 1 dV1(0+) di(0+) + = L R1 dt dtPutting corresponding values we get
dV1(0+) di(t)(0+) V1(0+) R1
= - dt dt L
Applying KCL at node ‘V2’1 V2 dV2
(V2 – V1)dt + + C = 0 … (ii) L R2 dtAt t = 0+(V2(0+) – V1(0+)) 1 dV2(0+) dV2(0+) + + C = 0 L R2 dt dtPutting corresponding values we get
dV2(0+) = 0 V/sec. dt
Equivalent circuit at t = 0+ V1(0+) V2(0+)
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Q#5.22: In the network shown in the figure, the switch K is closed at the instant t = 0, connecting an unenergized system to a voltage source. Show that if V(0) = V, then:di1(0+)/dt, di2(0+)/dt =? L1 L2
R1
L3
R3
i1 i2
R2
iL1(0-) = iL1(0+) = 0 AiL2(0-) = iL2(0+) = 0 AFor t 0According to KVLLoop 1:
From (ii) & (iv) we can determine the values of di1(0+)/dt & di2(0+)/dt.MASHAALLAH BHAI’S REFERENCE:
In order to indicate the physical relationship of the coils and, therefore, simplify the sign convention for the mutual terms, we employ what is commonly called the dot convention. Dots are placed beside each coil so that if currents are entering both dotted terminals or leaving both dotted terminals, the fluxes produced by these currents will add. If one current enters a dot and the other current leaves a dot, the mutual induced voltage and self-induced voltage terms will have opposite signs.
Q#5.23: For the network of the figure, show that if K is closed at t = 0, d2i1(0+)/dt2?CIRCUIT DIAGRAM:
L
R1 + -
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i1 C i2 R2
V(t)
Initial conditions:iL(0-) = iL(0+) = 0 A = i2(0+)VC(0-) = VC(0+) = 0 VEquivalent circuit after switching
V(t)i1(t) = R1
At t = 0+
V(0+)i1(0+) = R1
For t 0Loop 1: 1R1i1 + (i1 – i2)dt = V(t) … (i) CDifferentiating (i) with respect to ‘t’
di1 (i1 – i2) dV(t)R1 + = … (ii) dt C dt
+ -
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At t = 0+ di1(0+) (i1(0+) – i2(0+)) dV(0+)R1 + =
dt C dtPutting corresponding values we get
di1(0+) dV(0) V(0) 1= -
dt dt R1C R1
Differentiating (ii) with respect to ‘t’ d2i1 1 di1 di2 d2V(t)R1 + - =
dt2 C dt dt dt2
At t = 0+
d2i1(0+) 1 di1(0+) di2(0+) d2V(0+)R1 + - =
dt2 C dt dt dt2
From here we can determine the value of d2i1(0+)/dt2.
Q#5.24: The given network consists of two coupled coils and a capacitor. At t = 0, the switch K is closed connecting a generator of voltage, V(t) = V sin (t/(MC)1/2). Show that Va(0+) = 0, dVa(0+)/dt = (V/L)(M/C)1/2
, and d2Va(0+)/dt2 = 0.
CIRCUIT DIAGRAM:
M
K
a
Va
V(t)
+ -
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Equivalent circuit at t = 0+
After simplification
diL(0+)Va(0+) = VC(0+) + M dtWe know for t 0, according to KVL di 1L + idt = V(t) … (a) dt CAt t = 0+ di(0+) L + VC(0+) = V(0+) dt
+ -
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HereVC(0+) = 0 VV(t) = V sin (t/(MC)1/2)V(0+) = V sin ((0+)/(MC)1/2)V(0+) = V sin (0)V(0+) = V (0)V(0+) = 0 VPutting corresponding values we getdi(0+) = 0 Amp/sec. dt iL(0-) = iL(0+) = i(0+) = 0 ANow di(0+)Va(0+) = VC(0+) + M … (i) dtPutting corresponding values we getVa(0+) = 0 VoltNowDifferentiating (i) with respect to ‘t’dVa dVC d2i = + M … (b) dt dt dt2
Differentiating (a) with respect to ‘t’ d2i i dV(t)L + = … (a) dt2 C dtHere Vd(V(t)) = cos (t/(MC)1/2) (MC)1/2
At t = 0+ d2i(0+) i(0+) dV(0+)L + = … (c) dt2 C dtPutting corresponding values we getd2i(0+) V
= dt2 L(MC)1/2
At node a, apply KCL1 dVC
(VC – V(t)) + C = 0 … (c)L dtRearranging dVC(0+)iL(0+) + C = 0
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dt dVC(0+)0 + C = 0 dt
dVC(0+) = 0 dt
dVa dVC d2i = + M … (b) dt dt dt2
At t = 0+dVa(0+) dVC(0+) d2i(0+) = + M … (b) dt dt dt2
Putting corresponding values we get
dVa(0+) V M = dt L C
Differentiating (b) with respect to ‘t’d2Va d2VC d3i = + M dt2 dt2 dt3
Differentiating (c) with respect to ‘t’ d3i(0+) 1 di(0+) d2V(0+)L + = … (c) dt3 C dt dt2
d2V(0+) =? dt2
Vd(V(t)) = cos (t/(MC)1/2) (MC)1/2
-Vd2(V(t)) = sin (t/(MC)1/2) (MC)
-Vd2(V(0+)) = sin (0+/(MC)1/2) (MC) -V
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d2(V(0+)) = sin (0) (MC) -Vd2(V(0+)) = (0) (MC)
d2V(0+) = 0 dt2
d3i(0+) = 0 Amp/sec3 dt3 Differentiating (c) with respect to ‘t’(VC – V(t)) d2VC
+ C = 0 … (c) L dt2
At t = 0+(VC(0+) – V(0+)) d2VC(0+) + C = 0 … (c) L dt2
Putting corresponding values
d2VC(0+)= 0 V/sec2
dt2
d2Va d2VC d3i = + M dt2 dt2 dt3
At t = 0+d2Va(0+) d2VC(0+) d3i(0+) = + M dt2 dt2 dt3
d2Va(0+)= 0 Volt/sec2
dt2
Q#5.25: In the network of the figure, the switch K is opened at t = 0 after the network has attained a steady state with the switch closed. (a) Find an expression for the voltage across the switch at t = 0+. (b) If the parameters are adjusted such that i(0+) = 1 and di/dt (0+) = -1, what is the value of the derivative of the voltage across the switch, dVK/dt (0+) ?CIRCUIT DIAGRAM:
+ VK -
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R2
R1 C V
i
L
Initial conditions:i(0+) = 1di(0+) = -1 dt
Equivalent network after switching:
VK
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sc
V iL = R2
ViL(0-) = iL(0+) = R2
At t = 0+VK(0+) = VR1(0+)VR1(0+) = iL(0+)(R1)Putting corresponding value we get
VVR1(0+) = R1
R2
For t 0 1VK = iR1 + idt CDifferentiating with respect to ‘t’dVK di i = R1 +
dt dt CAt t = 0+dVK(0+) di(0+) i(0+) = R1 +
dt dt CPutting corresponding value we get
dVK(0+) 1 = - R dt C
Q#5.26: In the network shown in the figure, the switch K is closed at t = 0 connecting the battery with an unenergized system.
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(a) Find the voltage Va at t = 0+.(b) Find the voltage across capacitor C1 at t = .
CIRCUIT DIAGRAM:
R1
Va
K C1 C2
R2
V L
Equivalent network at t = 0+
After simplification:
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R2
Va(0+) = VEquivalent network at t =
VC1() = V.
Q#5.27: In the network of the figure, the switch K is closed at t = 0. At t = 0-, all capacitor voltages and inductor currents are zero. Three node to datum voltages are identified as V1, V2, and V3.
(a) Find V1 and dV1/dt at t = 0+.(b) Find V2 and dV2/dt at t = 0+.(c) Find V3 and dV3/dt at t = 0+.
CIRCUIT DIAGRAM:
V1 V3
+ -
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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} 66
V1(0+) = VC2(0-) = V – VR1(0-)Here VR1
VR1(0-) = R1 + R2 + R3
VR2 + VR3
V – VR1(0-) = R1 + R2 + R3
Q#5.29: In the network of the accompanying figure, a steady state is reached with the switch K closed and with i = I0, a constant. At t = 0, switch K, is opened. Find:
(a) V2(0-) =?(b) V2(0+) =?(c) (dV2/dt)(0+).
CIRCUIT DIAGRAM: 1 2
R2 +V2
C
R1 R3 L
-
Equivalent network at t = 0-
I0
I0
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After simplification we get
R1 R2
NowAccording to current divider rule: R1I0
iR2 = R1 + R2
We know V2(0-) = VL(0-) = 0Equivalent network at t = 0+
I0
I0
+ -
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R1
I0R1 VC(0+)
iL(0+)
After simplification
V2(0+)
VC(0+)
I0R1
+ -
+ -
+ -
+ -
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At node ‘V1’
V1 d(V1 – V2) + C = I0 … (i)R1 dt
At node ‘V2’
V2 d(V2 – V1) 1 + C + V2dt … (ii)R3 dt L
From eq. (ii)
d(V1 – V2) V2 1C = + V2dt dt R3 L
Substituting the value of Cd(V1 – V2)/dt in (i) we getV2 1 V1
+ V2dt + = I0
R3 L R1
At t = 0+
V2(0+) 1 V1(0+) + V2(0+)dt + = I0 … (iii) R3 L R1
Putting corresponding values we get I0R1R2
V1(0+) = R1 + R2
Differentiating eq. (iii) with respect to ‘t’ and from here putting the value of dV1(0+)/dt in eq. (i) we get dV2(0+)/dt.Hint:In eq. (i)V1(0+) I0R1R2
= R1 (R1 + R2)R1
dV2(0+) -I0R1R3
= dt C(R1 + R2)(R1 + R3)
dV1(0+) -R1 dV2(0+)=
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