Written Homework 3
PHYS 142 (Summer 2014)
Due: 7/29/2014
SOLUTION
RLC Circuit
Five infinite impedance voltmeters which
read rms values V1 , . . . , V5 are connected as
shown. The voltage source has constant amplitude V and adjustable angular frequency ω.
An alternating current with amplitude I flows
through the main body of the circuit. The reisistor has resistance R, the inductor has inductance L, and the capacitor has capacitance C.
Hint: Any function f (x) attains its maximum only
df
when dx
= 0.
For parts (b)-(d), give all answers in terms of R, L, and C.
a) Show that the current amplitude (as a function of ω) is I = p
V
. (3 pts)
+ (ωL − 1/ωC)2
p
V
Using that Z = R2 + (ωL − 1/ωC)2 and V = IZ, we have that I = p
.
2
R + (ωL − 1/ωC)2
R2
b) At what value of ω is the rms value V1 at a maximum? (3 pts)
Across a resistor, we have that V = IR, so that V will be at a maximum when I is at a maximum. Of
course, this occurs when Z is at a minimum, i.e. at resonance, so that
1
ω=√
.
LC
√
Finally, since Vrms = V / 2, Vrms also has its maximum for this value of ω.
c) At what value of ω is the rms value V2 at a maximum? (3 pts)
Again, Vrms has a maximum exactly when V does, and across an inductor V = IXL = IωL. Hence,
we compute
d
d V ωL
p
V (ω) =
dω
dω
R2 + (ωL − 1/ωC)2
VL
1
V ωL
1 1 =p
−
·
2
ωL
−
·
L
+
ωC
Cω 2
R2 + (ωL − 1/ωC)2 2 (R2 + (ωL − 1/ωC)2 )3/2
V
L 2
2
2L =
R
+
−
.
− 1/ωC)2 )3/2
ω2C 2
C
(R2 +
(ωL
0=
1
Solving for ω yields ω = p
.
LC − R2 C 2 /2
d) At what value of ω is the rms value V3 at a maximum? (3 pts)
As before we only need maximize V = IXC .
d
d V /ωC
p
V (ω) =
dω
dω
R2 + (ωL − 1/ωC)2
V
1
1 V
1 =− 2
+
−
·
2
ωL
−
·
L
+
2 ωC(R2 + (ωL − 1/ωC)2 )3/2
ωC
Cω 2
ω C(R2 + (ωL − 1/ωC)2 )1/2
2L −V((((((( R2
2
(
=
+
2ωL
−
.
(
(
2+
ωC
(((
ωC(R
(ωL − 1/ωC)2 )3/2 ω
(
r
1
R2
Solving for ω gives ω =
−
.
LC
2L2
0=
e) If R = 200 Ω, L = 0.40 H, C = 6.0 µF, V = 30 V, and ω = 200 rad/s, what are the readings of
the five voltmeters V1 , . . . , V5 ? (8 pts)
First we compute XL =√80 Ω and XC = 833 Ω. We then compute Z = 779 Ω, and from this arrive
at Irms = Vrms /Z = V / 2Z = 27 mA. Using Irms we can compute the rms voltages read by the first
three meters —
V1 = Irms R = 5.4 V
V2 = Irms XL = 2.2 V
V3 = Irms XC = 22.7 V.
For the remaining meters, we note that across any two points x and y, we always have that Vxy = IZxy ,
where Zxy is the total impedance between those two points. A similar statement holds for the rms values.
Since the total impdeance across the terminals of V4 is Zcb = |XL − XC | = 753 Ω, we have
V4 = Irms Zcb = 20.5 V.
Finally, we have that
V5 = Irms Z = 21.2 V.
TOTAL (of 20)