Physics 212 Lecture 21, Slide 1 Physics 212 Lecture 21 Resonance and power in AC circuits
Physics 212 Lecture 21, Slide 1
Physics 212 Lecture 21
Resonance and power in AC circuits
Physics 212 Lecture 21, Slide 2
Imax XL
Imax XC
Imax R
L
R
C
emax
Imax R
emax = Imax Z
Imax(XL-XC)
Imax XL
Imax XC
Imax R
w
XL = wL
XC = 1/wC
R
f XL-XC
Physics 212 Lecture 21, Slide 3
Peak AC Problems
• “Ohms” Law for each element – PEAK values
– Vgen = Imax Z
– VResistor = Imax R
– Vinductor = Imax XL
– VCapacitor = Imax XC
• Problem A generator with peak voltage 15 volts and angular
frequency 25 rad/sec is connected in series with an 8 Henry inductor, a 0.4 mF capacitor and a 50 ohm resistor. What is the peak current through the circuit?
07
L
R
C
200 LX Lw
1100 CX Cw
22 112 L CZ R X X
AZ
VI
gen13.0max
XL
XC
R
Physics 212 Lecture 21, Slide 5
Peak AC Problems
• “Ohms” Law for each element
– Vgen = I Z
– VResistor = I R
– Vinductor = I XL
– VCapacitor = I XC
• Typical Problem A generator with peak voltage 15 volts and angular frequency 25 rad/sec is
connected in series with an 8 Henry inductor, a 0.004 Farad capacitor and a 50 ohm resistor. What is the peak current through the circuit?
Which element has the largest peak voltage across it?
A) Generator E) All the same.
B) Inductor
C) Resistor
D) Capacitor
max maxV I X
12
L
R
C
LX Lw
22
L CZ R X X
1CX Cw
200 LX Lw
1100 CX Cw
22 112 L CZ R X X
AZ
VI
gen13.0max
XL
XC
R
Physics 212 Lecture 21, Slide 6
Peak AC Problems
• “Ohms” Law for each element
– Vgen = I Z
– VResistor = I R
– Vinductor = I XL
– VCapacitor = I XC
• Typical Problem A generator with peak voltage 15 volts and angular frequency 25 rad/sec is
connected in series with an 8 Henry inductor, a 0.4 mF capacitor and a 50 ohm resistor. What is the peak current through the circuit?
What happens to the impedance if we decrease the angular frequency to 20 rad/sec?
A) Z increases
B) Z remains the same
C) Z decreases (XL-XC): (200-100) (160-125)
14
L
R
C
LX Lw
22
L CZ R X X
1CX Cw
XL
XC
R
XL
XC
R
Z25 Z20
Physics 212 Lecture 21, Slide 7
Resonance
Light-bulb Demo
Physics 212 Lecture 21, Slide 8
Resonance
R is independent of w
XL increases with w
LLX w
is minimum at resonance
22 )( CL XXRZ
XC increases with 1/w 1
CCXw
Resonance: XL = XC 0
1
LCw
10
Frequency at which voltage across inductor and capacitor cancel
Resonance in AC Circuits
frequency
Imp
ed
an
ce
R XC
Z
XL w0
Z = R at resonance
Physics 212 Lecture 21, Slide 9
Imax XL
Imax XC
Imax R
Case 1
Imax XL
Imax XC
Imax R
Case 2
Resonance: XL = XC
Z = R
Same since R doesn't change
Checkpoint 1a
Consider two RLC circuits with identical generators and resistors. Both circuits
are driven at the resonant frequency. Circuit II has twice the inductance and 1/2
the capacitance of circuit I as shown above.
Compare the peak voltage across the resistor in the two circuits
A. VI > VII B. VI = VII C. VI < VII
Physics 212 Lecture 21, Slide 10
Voltage in second circuit will be twice that of the first because of the 2L compared to L
Checkpoint 1b
Consider two RLC circuits with identical generators and resistors. Both circuits
are driven at the resonant frequency. Circuit II has twice the inductance and 1/2
the capacitance of circuit I as shown above.
Compare the peak voltage across the inductor in the two circuits
A. VI > VII B. VI = VII C. VI < VII Imax XL
Imax XC
Imax R
Case 1
Imax XL
Imax XC
Imax R
Case 2
Physics 212 Lecture 21, Slide 11
The peak voltage will be greater in circuit 2 because the value of XC doubles.
Checkpoint 1c
Consider two RLC circuits with identical generators and resistors. Both circuits
are driven at the resonant frequency. Circuit II has twice the inductance and 1/2
the capacitance of circuit I as shown above.
Compare the peak voltage across the capacitor in the two circuits
A. VI > VII B. VI = VII C. VI < VII Imax XL
Imax XC
Imax R
Case 1
Imax XL
Imax XC
Imax R
Case 2
Physics 212 Lecture 21, Slide 12
The voltage across the inductor and the capacitor are equal when at resonant frequency, so there is no lag or lead.
Checkpoint 1d
Consider two RLC circuits with identical generators and resistors. Both circuits
are driven at the resonant frequency. Circuit II has twice the inductance and 1/2
the capacitance of circuit I as shown above.
At the resonant frequency, which of the following is true?
A. Current leads voltage across the generator
B. Current lags voltage across the generator
C. Current is in phase with voltage across the generator
Imax XL
Imax XC
Imax R
Case 1
Imax XL
Imax XC
Imax R
Case 2
Physics 212 Lecture 21, Slide 13
Z
Quality factor Q
In general
U
UQ
max2
Umax = max energy stored U = energy dissipated in one cycle at resonance
Physics 212 Lecture 21, Slide 14
Larger Q, sharper resonance
Physics 212 Lecture 21, Slide 15
Series LCR circuit
1. The current leads generator voltage by 45o.
2. Assume V = Vmaxsinwt What does the phasor diagram look like at t = 0?
V = Vmax sinwt V is horizontal at t = 0 (V = 0)
C
R
L V ~
(A) (B) (C) (D) X X X
L C RV V V V VL < VC if current leads generator voltage
Physics 212 Lecture 21, Slide 16
The current leads generator voltage by 45o
C
R
L V ~
(A) decrease w (B) increase w (C) Not enough info
Original w
f
Increase w
At resonance (w0) At resonance
XL = XC
XL increases
XC decreases
How should we change w to bring circuit to resonance?
Series LCR circuit
Physics 212 Lecture 21, Slide 17
1. The current leads generator voltage by 45o.
2. Suppose XC = 2XL at frequency w.
C
R
L V ~
f
By what factor should we increase w to bring circuit to resonance w0 ?
2C LX X2 2
0
1 1 12
2 2L
C LCw w w
w
02
w
w
02 2
w
w
02
w
w
04
w
w(A) (B) (C) (D)
Series LCR circuit
Physics 212 Lecture 21, Slide 18
Consider the harmonically driven series LCR circuit shown. At some frequency w, XC = 2XL = R Vmax = 100 V R = 50 k
C
R
L V ~
At resonance XL = XC Z R
maxmax 0 3
100( ) 2 mA
50 10
VI
R xw
If we change w, what is the maximum current that can flow?
maxI 2 mA maxI 2 mA maxI 2 2 mA (A) (B) (C)
Series LCR circuit
Physics 212 Lecture 21, Slide 19
L
R
C
0
1T
P t P t dtT
Average, over one cycle, of any periodic function :
P(t) varies in time. We want the average power.
Instantaneous power
P(t) = I(t)V(t)
For harmonically-varying functions like sin wt or cos wt , T = 2π/ω .
0
1T
P t V t I t dtT
Circuit element:
,
0
1sin cos 0
T
L CP t t t dtT
w w For L or C, voltage and
current are /2 out of phase:
I sin wt, V cost wt
22 2
0
1 1sin
2
T
R peak peakP t I R I t Rdt I RT
w
Average power for R is not zero since V and I are in phase.
Define root mean square (RMS)
2 2 21
2peak RMSI R I R I R
1.707
2RMS peak peakI I I
1 1cos
2 2
peak
peak peak peakI R IZ
ee f
R
f XL-XC Z
Average power dissipated in R for LCR circuit
Maximum power delivered when f = 0 … at resonance !
1cos
2peak peakPower I e f
R
f XL-XC Z
Imax w0L
Imax / w0C
Imax R = emax
w0L = 1/w0C
f = 0
At resonance,
0 max 0
max
C L
R R
LI LV VQ
V V RI R
w w
Physics 212 Lecture 21, Slide 22
Power Transmission
• If you want to deliver 1500 Watts at 100 Volts over transmission lines w/ resistance of 5 Ohms. How much power is lost in the lines?
– Current Delivered: I = P/V = 15 Amps
– Loss from power line = I2 R = 15*15 * 5 = 1,125 Watts!
• If you deliver 1500 Watts at 10,000 Volts over the same transmission lines. How much power is lost?
– Current Delivered: I = P/V = 0.15 Amps
– Loss from power line = I2R = 0.15*0.15*5 = 0.113 Watts
DEMO
Physics 212 Lecture 21, Slide 23
Transformer
Iron core maintains same B throughout.
Primary coil captures flux P = NP B
Secondary coil captures flux S = NS B
Faraday’s Law:
VS = d S /dt = NS B
VP = d P /dt = NP B p p
s s
V N
V N
How do we transform 10,000 V into 120 V for your toaster ?
Time-varying voltage in primary causes
time-varying B in iron-core.
Flux per turn = B (Area) = B
7200 V / 240 V