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Lecture 30
AC power. Resonance.
Transformers.
ACT: εC circuit
A capacitor is connected to an AC generator as
shown. If the frequency of the generator is

C

doubled, the amplitude of the current in the
circuit will

ε

A. increase by a factor of 2
B. not change
C. decrease by a factor of 2

I =

E
= EC ω
Z

Z =

(X

L

− XC

)

2

+ R 2 = XC =

1
Cω
Filters
Vout

Vout depends on frequency:
High ω ⇒ smaller reactance ⇒ VC = Vout → 0

C

Low ω ⇒ larger reactance ⇒ no current flows
through R ⇒ smaller VR ⇒ VC = Vout → ε
VL
ε

This is a circuit that only passes
low frequencies: low-pass filter
 Bass knob on radio
ω0 = 1

RC

R

ε

ω

If instead we look at the voltage through the resistor: high-pass filter
 Treble control
More filters
L

High ω ⇒ large XL ⇒ VL ~ ε ⇒

R

High-pass filter is Vout = VL
Low-pass filter is Vout = VR

ε
L
R

C

ε

VR ~ 0 and I ~ 0

VL
ε

Low ω → I ~ 0 due to capacitor
High ω → I ~ 0 due to inductor
Band pass filter (back to this
later: resonance)

ω0 = 1

RC

Vout
ε

ω0

ω
ACT: εRL circuit
An RL circuit is driven by an AC generator
as shown in the figure. The current

through the resistor and the generator
voltage are:
A. always out of phase
B. always in phase

C. sometimes in phase and sometimes out
of phase

And this is the current through all
elements, by the way…

E
VL

I
VR
ACT: εRC circuit
A series RC circuit is driven by emf ε =E sinωt. Which of the
following could be an appropriate phasor diagram?
VL

VC

E

E
VC

VR
VR

VR

A

B

~

VC

C

For this circuit, which of the following is true?
(a) The drive voltage is in phase with the current.
(b) The drive voltage lags the current.
(c) The drive voltage leads the current.

E
ACT: Bring in phase
The current and driving voltage in an RLC circuit are shown in the graph.
How should the frequency of the power source be changed to bring these
two quantities in phase?

A. Increase ω
B. Decrease ω
C. Current and driving voltage
cannot be in phase.

0

i

ε

t

From the figure, current leads driving voltage ⇒ ϕ < 0
⇒ XC > XL ⇒ to make them equal, frequency needs to
increase.

IXL
IR
IXC

ε
ACT: Bring in phase
The current and driving voltage in an RLC circuit are shown in the graph.
Which of the following phasor diagrams represents the current at t = 0?

A.

B.

I
I

t
i (t) is the horizontal projection of the phasor.
I

C.

0

i

ε

From the figure:
At t = 0, i ~ 2/3I (>0)
And it should be increasing.
Resonance
E

Current amplitude in a series RLC circuit driven I =
Z
by a source of amplitude E :
Maximum current when impendance Z =
i.e., when X L = XC
1
Lω =
Cω

ω0 =

E
R

I =
I

0

0

1

LC

(X

L

− XC

)

2

+ R 2 is minimum

Resonance:
Driving frequency =
natural frequency

E
E
= cos ϕ
Z
R

Maximum current ⇔ maximum cosϕ ⇔
cosϕ ~ 1 ⇔ ϕ ~ 0 (circuit in phase)
ω

2ωo

It’s the bandpass filter!
ACT: Resonance

This circuit is being driven __________
its resonance frequency.
A. above
B. below
C. exactly at

To achieve resonance, we need to decrease XL and to
increase XC ⇒ to decrease frequency ω
Power in AC circuits
Instantaneous power supplied to the circuit:

(

P (t ) ≡ ε (t ) i (t ) = ( E cos ω ) I cos ( ω − ϕ )
t
t

Often more useful: Average power

P (t ) = EI cos ω cos ( ω −ϕ)
t
t

1
P = EI cos ϕ
2

cos ω cos ( ω −ϕ) = cos ω ( cos ω cos ϕ − sin ω sin ϕ)
t
t
t
t
t
1 2π
1
2
∫0 cos xdx = 2
2π
1 2π
cos x sin x =
∫0 cos x sin xdx = 0
2π
cos2 x

Define:

E rms =

E
2

=

I rms =

I
2

P = E rms Irms cos ϕ

)
Power factor
P = E rms Irms cos ϕ

Power factor (PF)

Maximum power ⇔ ϕ = 0 ⇔

I =

Resonant circuit

E
E
=
cos ϕ
Z
R
2
P = I rms R

All energy dissipation happens
at the resistor(s).
The Q factor
How “sharp” is the resonance? (ie, the resonance peak)

Umax
Q ≡ 2π
∆U
• For RLC circuit, Umax =

Umax is max energy stored in the system
ΔU is the energy dissipated in one cycle

1
LI 2
2

period

1 2
1 2  2π
• Losses only come from R : ∆ U = I RT = I R 
ω
2
2
 0

ωL
0

Q =
ω0 =

1

LC

Q =

R

1

X L ,0
=
R

ωCR
0

=

XC ,0
R


÷
÷


X L ,0
XC ,0
Q =
=
R
R
X L ,0
XC ,0
Q =
=
R
R

Large Q ⇒ sharp peak ⇒ better “quality”

L and C control how much energy is stored.
R controls how much energy is lost.

Small resistance ⇒ Large Q
Transformers
Application of Faraday’s Law
• Changing EMF in primary coil creates changing flux
• Changing flux creates EMF in secondary coil.
V =N

Magnetic flux remains
mostly in the core.
Core “directs” B lines

d ΦB
for both coils
dt

d ΦB V1
V
=
= 2
dt
N1 N2

V1 V2
=
N1 N2

V1

Efficient method to change
voltage for AC.

N1

N2

d ΦB
dt

If no energy is lost in the coils, power on both sides must be the same

V1I1 = V2I2

V2
In-class example: Jacob’s ladder
A transformer outputs Vrms = 20,000 V when it is plugged into a wall
source (Vrms = 120 V). If the primary coil (coil hooked to the wall)
has 1667 loops, how many loops does the secondary coil have?
A. 10
B. 278
C. 1667
D. 10,000

V1 V2
=
N1 N2
N2 = N1

V2
20, 000 V
= ( 1667 )
= 278, 000
V1
120 V

E. 278,000
DEMO:
Jacob’s ladder

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AC Circuits and Resonance

  • 1. Lecture 30 AC power. Resonance. Transformers.
  • 2. ACT: εC circuit A capacitor is connected to an AC generator as shown. If the frequency of the generator is C doubled, the amplitude of the current in the circuit will ε A. increase by a factor of 2 B. not change C. decrease by a factor of 2 I = E = EC ω Z Z = (X L − XC ) 2 + R 2 = XC = 1 Cω
  • 3. Filters Vout Vout depends on frequency: High ω ⇒ smaller reactance ⇒ VC = Vout → 0 C Low ω ⇒ larger reactance ⇒ no current flows through R ⇒ smaller VR ⇒ VC = Vout → ε VL ε This is a circuit that only passes low frequencies: low-pass filter  Bass knob on radio ω0 = 1 RC R ε ω If instead we look at the voltage through the resistor: high-pass filter  Treble control
  • 4. More filters L High ω ⇒ large XL ⇒ VL ~ ε ⇒ R High-pass filter is Vout = VL Low-pass filter is Vout = VR ε L R C ε VR ~ 0 and I ~ 0 VL ε Low ω → I ~ 0 due to capacitor High ω → I ~ 0 due to inductor Band pass filter (back to this later: resonance) ω0 = 1 RC Vout ε ω0 ω
  • 5. ACT: εRL circuit An RL circuit is driven by an AC generator as shown in the figure. The current through the resistor and the generator voltage are: A. always out of phase B. always in phase C. sometimes in phase and sometimes out of phase And this is the current through all elements, by the way… E VL I VR
  • 6. ACT: εRC circuit A series RC circuit is driven by emf ε =E sinωt. Which of the following could be an appropriate phasor diagram? VL VC E E VC VR VR VR A B ~ VC C For this circuit, which of the following is true? (a) The drive voltage is in phase with the current. (b) The drive voltage lags the current. (c) The drive voltage leads the current. E
  • 7. ACT: Bring in phase The current and driving voltage in an RLC circuit are shown in the graph. How should the frequency of the power source be changed to bring these two quantities in phase? A. Increase ω B. Decrease ω C. Current and driving voltage cannot be in phase. 0 i ε t From the figure, current leads driving voltage ⇒ ϕ < 0 ⇒ XC > XL ⇒ to make them equal, frequency needs to increase. IXL IR IXC ε
  • 8. ACT: Bring in phase The current and driving voltage in an RLC circuit are shown in the graph. Which of the following phasor diagrams represents the current at t = 0? A. B. I I t i (t) is the horizontal projection of the phasor. I C. 0 i ε From the figure: At t = 0, i ~ 2/3I (>0) And it should be increasing.
  • 9. Resonance E Current amplitude in a series RLC circuit driven I = Z by a source of amplitude E : Maximum current when impendance Z = i.e., when X L = XC 1 Lω = Cω ω0 = E R I = I 0 0 1 LC (X L − XC ) 2 + R 2 is minimum Resonance: Driving frequency = natural frequency E E = cos ϕ Z R Maximum current ⇔ maximum cosϕ ⇔ cosϕ ~ 1 ⇔ ϕ ~ 0 (circuit in phase) ω 2ωo It’s the bandpass filter!
  • 10. ACT: Resonance This circuit is being driven __________ its resonance frequency. A. above B. below C. exactly at To achieve resonance, we need to decrease XL and to increase XC ⇒ to decrease frequency ω
  • 11. Power in AC circuits Instantaneous power supplied to the circuit: ( P (t ) ≡ ε (t ) i (t ) = ( E cos ω ) I cos ( ω − ϕ ) t t Often more useful: Average power P (t ) = EI cos ω cos ( ω −ϕ) t t 1 P = EI cos ϕ 2 cos ω cos ( ω −ϕ) = cos ω ( cos ω cos ϕ − sin ω sin ϕ) t t t t t 1 2π 1 2 ∫0 cos xdx = 2 2π 1 2π cos x sin x = ∫0 cos x sin xdx = 0 2π cos2 x Define: E rms = E 2 = I rms = I 2 P = E rms Irms cos ϕ )
  • 12. Power factor P = E rms Irms cos ϕ Power factor (PF) Maximum power ⇔ ϕ = 0 ⇔ I = Resonant circuit E E = cos ϕ Z R 2 P = I rms R All energy dissipation happens at the resistor(s).
  • 13. The Q factor How “sharp” is the resonance? (ie, the resonance peak) Umax Q ≡ 2π ∆U • For RLC circuit, Umax = Umax is max energy stored in the system ΔU is the energy dissipated in one cycle 1 LI 2 2 period 1 2 1 2  2π • Losses only come from R : ∆ U = I RT = I R  ω 2 2  0 ωL 0 Q = ω0 = 1 LC Q = R 1 X L ,0 = R ωCR 0 = XC ,0 R  ÷ ÷  X L ,0 XC ,0 Q = = R R
  • 14. X L ,0 XC ,0 Q = = R R Large Q ⇒ sharp peak ⇒ better “quality” L and C control how much energy is stored. R controls how much energy is lost. Small resistance ⇒ Large Q
  • 15. Transformers Application of Faraday’s Law • Changing EMF in primary coil creates changing flux • Changing flux creates EMF in secondary coil. V =N Magnetic flux remains mostly in the core. Core “directs” B lines d ΦB for both coils dt d ΦB V1 V = = 2 dt N1 N2 V1 V2 = N1 N2 V1 Efficient method to change voltage for AC. N1 N2 d ΦB dt If no energy is lost in the coils, power on both sides must be the same V1I1 = V2I2 V2
  • 16. In-class example: Jacob’s ladder A transformer outputs Vrms = 20,000 V when it is plugged into a wall source (Vrms = 120 V). If the primary coil (coil hooked to the wall) has 1667 loops, how many loops does the secondary coil have? A. 10 B. 278 C. 1667 D. 10,000 V1 V2 = N1 N2 N2 = N1 V2 20, 000 V = ( 1667 ) = 278, 000 V1 120 V E. 278,000 DEMO: Jacob’s ladder