Discussion Question 13A
P212, Week 13
Power in AC Circuits
An electronic device, consisting of a simple RLC circuit, is designed to be connected to an
American-standard power outlet delivering an EMF of 120 Vrms at 60 Hz frequency. However,
you have transported this device to Europe, where standard AC power outlets deliver a blazing
240 Vrms. To overcome this problem, you connect your device to the European wall outlet
through a step-down transformer with Np turns in the primary coil and Ns turns in the secondary.
The values of the resistance, capacitance, and inductance in your device are given in the figure.
Ep
R
Np
Ns
C
L
Ep,rms = 240 V
Erms = 120 V
f = 60 Hz
R = 3.8 Ω
L = 0.05 H
C = 177 µF
(a) First, let’s design the transformer. What relation must you impose on Np and Ns to drop the
European line voltage Ep down to E = 120 Vrms?
Try thinking physically: As is typical, your transformer has an iron core which ensures that
the magnetic field produced in the primary coil is transferred exactly to the secondary coil.
Now E = -dΦ/dt … to step down the voltage, which coil needs to have more turns?
The field in the primary will be the same as in the secondary . The flux
on the primary will be Φ p = BN P A. The flux on the secondary will be Φ s = BN s A.
By Faraday's law: E p = −
We want
E
dΦ
dB
dB
N
= −NP A
→ s = s
; Es = − N s A
Ep N p
dt
dt
dt
1
N
Es 1
= → s =
Np 2
Ep 2
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(b) Let the time-dependent EMF produced by the Euro generator be Ep(t) = Ep,max sin(ωt) which
(
)
means E ( t ) = 120 2 sin ωt where E ( t ) EMF delivered to the secondary side of the
transformer . What is the time-dependent current I(t) produced in the load? Write your answer
in the form I(t) = Imax sin(ωt-φ), and determine the numerical values of Imax and φ.
We begin by computing the reactances:
ω = 2π (60) = 377 rad/s
X L = ω L = 377 ( 0.05 ) = 18.85Ω
XC =
3.80Ω
1
1
=
= 14.99Ω
ωC 377 (177 × 10−6 )
18.85Ω
14.99Ω
5.419Ω
3.865Ω
45.4º
3.80Ω
X = X L − X C = 3.86Ω
Ηere is a phasor diagram for reactances. Z = 5.419 φ = 45.40
I max = Emax / Z =
120 2
= 31.32 A Hence I=31.32 sin(ωt − 45.40 ) since E lags I
5.419
(c) What is the average power <P> delivered to the load?
Your formula sheet has a couple of “cookbook” formulas you can use.
P = Erms I rms cos φ = 120 ∗
31.4
∗ cos( 45.40 ) = 1.87 kW
2
(d) Obtain an algebraic expression for the <P> delivered to the load in terms of Erms, R, and
X ≡ XL - XC . Hint -write a phasor diagram for Z in terms of X and L and use trig to find the
required phase angle.
⎛E
P = Erms I rms cos φ = Erms ⎜ rms
⎝ Z
From triangle cosφ =
2
Erms
⎞
cos
φ
.
From
triangle
Z=R/cos
φ
hence
P
cos 2 φ
=
⎟
R
⎠
φ
X
2
⎞
R
R E2
E2 ⎛
→ P = rms ⎜
= 2 rms 2
⎟
R ⎝ R2 + X 2 ⎠
R +X
R2 + X 2
R
Z
R
(e) (e)
(e) Suppose your European AC generator has a variable frequency f, though the EMF it supplies
is fixed at 240 Vrms. What f would you select so that the maximum possible power is delivered to
your device? Hint- sketch the expression for <P> you obtained in part (d) as a function of X.
Which value of X maximizes <P>? What frequency gives you that value for X?
<P>
The formula says P monotonically decreases with X. So the maximal P
when X=0 or X L = X C → ω L =
1
2π
→ f =
= 53.5 Hz
ωC
LC
X
Page 2 of 4
is
(f) What is the maximum average power <P>max that you can deliver to your device?
2
2
R Erms
Erms
=
P
.
the
maximum
power
is
just
R2 + X 2
R
This makes sense since at resonance the load is purely resistive and this is
From our expression P =
the familar result for a resistor, . P =
2
Erms
1202
=
= 3.79 kW
3.8
R
Consider a slightly different situation. Let’s say you built your device to be purely inductive →
it contains nothing more than a coil you wound yourself, giving L = 0.05 H. However, no device
is perfect: every electronic circuit element contains some residual inductance, capacitance, and
resistance. You connect your device to the European wall outlet and measure the response in the
load. You find a current with an rms value of Irms = 8.5 A and lags the transformer EMF by 30°.
Ep
Ep,rms = 240 V
Erms = 120 V
f = 60 Hz
L = 0.05 H
Irms = 8.5 A
|φ| = 30º
R
Np
Ns
C
L
(g) What is the resistance R of your device?
You will need a couple of equations to “reverse engineer” this circuit: you measured the
voltage, current, and phase and now you have to figure out R. Big hint: obtain an equation
for R in terms of the impedance Z and the phase φ. (It's a useful equation, you might want to
remember it!)
From triangle Z = R / cos φ or R = Z cos φ . Z =
→R=
Erms
⎛ 120 ⎞
0
cos φ = ⎜
⎟ cos 30 = 12.23 Ω
8.5
I rms
⎝
⎠
Erms
I rms
Z
φ
X
R
(h) What is the capacitance C of your device?
From triangle X = Z sin φ =
→C =
1
⎛
ω ⎜ωL −
⎝
→ C = 225µ F
⎞
Erms
s in φ ⎟
I rms
⎠
=
Erms
1
s in φ =ω L −
I rms
ωC
Z
φ
1
120
⎛
⎞
( 2π × 60) ⎜ ( 2π × 60)0.05 −
s in 30o ⎟
8 .5
⎝
⎠
X
R
Page 3 of 4
(i) In part (c), you probably used the average-power formula <P>=Erms Irms cos(φ) from your
formula sheet (the others, like <P>=Irms2 R, are equivalent). Let’s derive that result! First, what
is the instantaneous power P(t) delivered to the load? Give an algebraic answer (no numbers) in
terms of Erms, Irms, φ, ω, and t.
Instantaneous power is P = IV, forever and ever, under all circumstances. Remember the
time-dependent expressions you found in part (b) … and in preparation for the next question,
you should “massage” your result using sin( a − b ) = sin a cos b − cos a sin b .
Let E = E max sin ωt and I = I max sin (ωt − φ ) where φ is the phase lag of
the current relative to the voltage. P =EI = E max I max sin ωt sin (ωt − φ )
→ P = E max I max sin ωt ( sin ωt cos φ − cos ωt sin φ )
→ P = E max I max ( cos φ sin 2 ωt − sin φ sin ωt cos ωt )
(j) Now calculate the time-average of your result for P(t) over one cycle to obtain the average
power <P> delivered to the load.
Remember, the average of sin2 and cos2 is 1/2, while the average of (sin⋅cos) is zero.
P = E max I max ( cos φ sin 2 ωt − sin φ sin ωt cos ωt )
(
→ P = E max I max cos φ sin 2 ωt − sin φ sin ωt cos ωt
1
)
As suggested by the figure sin 2 ωt (solid) averages to 1/2 and
sin ωt cos ωt (dashed curve) averages to zero over one cycle.
Hence P =
E max I max
cos φ = E rms I rms cos φ
2
0.5
0
0
1
2
3
4
5
-0.5
-1
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