Discussion Question 12C
P212, Week 12
RLC Circuits
R
(a) Calculate the maximum EMF Em
and the maximum current Im in the RLC
circuit described at right.
The “rms” = root-mean-square value of
anything oscillating sinusoidally is its
peak value divided by √2.
L
C
E
X L = ωL
The figure shows R, X L and
Em = Erms √2 = 170 V
Im = Em/Z = 170/223.6
= 0.76 A
X C . The net reactance is 100Ω
in the direction of capacitance.
R = 200 Ω
L = 40 mH
C = 0.20 μF
Erms = 120 V
ω = 104 rad/sec
= 400 Ω
200 Ω
R = 200 Ω
1
ωC
= 500 Ω
XC =
The right triangle gives
Z = 223.6 Ω.
100 Ω
Z=223.6 Ω
(b) Find the magnitude and sign of the phase φ by which the driving EMF leads the
current.
A negative phase means that the driving EMF lags the current … which is the case here?
Does your answer make sense given the reactances you calculated earlier?
200 Ω
26.60
Z=223.6 Ω
100 Ω
Here is our reactance triangle from before. Z ( ∝ E ) lags R
( ∝ I ) by φ = −Tan -1 ⎛⎜
100 ⎞
0
⎟ = −26.6
⎝ 200 ⎠
(c) Draw the phasor diagram for this
circuit, giving numerical values for the lengths of each phasor (E, VR, VC, VL).
Be sure to draw your diagram carefully: use longer phasors for larger peak voltages.
VR = Im R = 152 V
VC = Im / (ωC) = 380 V
VL = Im ω L = 304
VL = 400 × 0.76
VR = 200 × 0.76
VC = 500 × 0.76
R
(d) What is the resonant frequency ω0 of
this circuit?
L
C
ωo = 1/√(LC) = 11,200 rad/sec
R = 200 Ω
L = 40 mH
C = 0.20 μF
Erms = 120 V
ω = 104 rad/sec
E
(e) Calculate the maximum energies UL,max and UC,max stored in the inductor and capacitor.
UL,max = ½ L Im2 . Inserting Im= 0.76 A from part (a) we have UL,max = 11.6 mJ
UC,max = ½ C (VC,max)2 . Inserting VC,max = 380 V from part (c) gives us UC,max = 14.4 mJ
(f) Assume that the angular frequency ω of the generator is variable. For what ω is the
total impedance Z equal to 2R?
R 2 + ( X L − X C ) = 2 R → R 2 + ( X L − X C ) =4 R 2 → ( X L − X C ) = 3R 2
2
→ ( X L − XC )
Let ω0 =
2
2
2
2
2
1 ⎞
1 ⎞
⎛
⎛
⎛R⎞
2
= ⎜ωL −
⎟ = 3R → ⎜ ω −
⎟ = 3⎜ ⎟
ωC ⎠
ω LC ⎠
⎝
⎝
⎝L⎠
2
1
R
=11.18 kHz and ω1 = = 5 kHz
L
LC
2
⎛
ω02 ⎞
2
2
2 2
2 2
Then ⎜ ω −
⎟ = 3ω1 → (ω − ω0 ) = 3ω1 ω
ω ⎠
⎝
ω 4 − 2ω02ω 2 + ω04 = 3ω12ω 2 → ω 4 − ( 2ω02 + 3ω12 ) ω 2 + ω04 = 0
Using quadratic equation: ω 2 =
325 × 106 ±
( 325 ×10 )
6 2
( 2ω
2
0
+ 3ω12 ) ±
( 2ω
2
0
+ 3ω12 ) − 4ω04
2
2
− 4 (11.18 × 103 )
4
325 × 106 ± 207 ×106
2
2
2
6
6
ω = 266 ×10 and 59 ×10 → ω = 16.3 kHz and 7.66 kHz
ω2 =
=
2