Discussion Question 12A
P212, Week 12
LC Circuits
At right, we see the classic LC circuit, consisting of an inductor in
series with a capacitor. To be precise, this is an undriven LC circuit,
since there is no battery driving the flow of current. Nevertheless,
L
this simple circuit has an amazing and highly useful property: it
supports a resonant oscillation of current. In the ideal case shown
here where there is no resistance in the circuit, the oscillation can
continue indefinitely without any external source of EMF to drive it. LC
circuits often appear as parts of larger networks which are designed to
operate at a particular frequency. The most familiar example is a radio,
whose circuitry can be tuned to receive (i.e. respond to) incoming
electromagnetic waves of a particular frequency.
C
L = 300 mH
C = 0.025 µF
(a) What is the angular frequency ω0 of the current maintained by the circuit? What is
the linear frequency f0?
An undriven LC circuit can only operate in a stable way at its natural resonant frequency. As
for linear versus angular frequency, the conversion is easy if you remember the units: f is in
Hz = 1/sec, ω is in rad/sec.
The natural frequency is when the net reactance is 0 or ω L −
ω=
1
=0→
ωC
1
1
=
= 11.55 kHz ;
−6
LC
0.3 ( 0.025 × 10 )
ω0 =2π f 0 → f 0 =
ω0 11.55 kHz
=
= 1838 Hz
2π
6.28318
(b) Let’s set our clock so that the current in the circuit is at its maximum value Imax at time
t = 0. Write down an expression for the time-dependence of the current I(t) in terms of Imax
and the frequency ω0.
It’s either a sine or a cosine …
We are looking for a trig function that maximizes as t = 0. Hence I = I max cos ωt
(c) Starting from your expression for I(t), determine the voltages VL(t)
and VC(t) across the inductor and capacitor. As part of your solution,
find the peak values VL,max and VC,max in terms of Imax.
You’ll need your familiar formulas for the voltage across R, C, and
L ... and remember that, by definition, I =
dQ
and so Q = ∫ I dt .
dt
L
C
L = 300 mH
C = 0.025 µF
dI
d
dωt ⎛ d cos ωt ⎞
= L ( I max cos ωt ) = LI max
⎜
⎟ = −ω L I max sin ωt → VL ,max = ω L I max
dt
dt
dt ⎝ dωt ⎠
dQ
VC = Q / C If the current is positive in the clockwise direction, I =
dt
I
I
I
Q = ∫ dt I = ∫ dt I max cos ωt = max sin ωt → VC = max sin ωt → VC,max = max
ω
ωC
ωC
VL = L
(d) To visualize what’s going on, sketch your functions I(t), VL(t), and VC(t). Don’t worry
about the amplitudes of your curves, just their shapes and phases.
t
π/2ω
VL
π/ω
I
3π/2ω
2π/ω
VC
2
(e) Have a look at your plot → does VL lead or lag the current? How about VC?
Leading and lagging can be tricky concepts. Think of it this way: which one “gets there”
(i.e. reaches its maximum value) first, V or I? The one that “gets there first” leads the other.
Another way of viewing leading and lagging is if
I ∝ cos ωt and V ∝ cos (ωt + δ ) , we say V leads I by δ
VL leads. VC lags. We can do this using cos (ωt + δ ) = cos ωt cosδ − sin ωt sinδ
π⎞
⎛
VL ∝ − sin ωt = cos ⎜ ωt + ⎟ hence VL leads I by 900
2⎠
⎝
π⎞
⎛
VC ∝ sin ωt = cos ⎜ ωt − ⎟ hence VC lags I by 900
2⎠
⎝
Congratulations! You’ve just derived the master relations between current and voltage for
inductors and capacitors in an AC circuit!
Peak Values
VR,max = Imax R
VL,max = Imax XL → XL = ωL
VC,max = Imax XC → XC = 1/ωC
Relative Phases
Across R → V in phase with I
Across L → V leads I by 90°
Across C → V lags I by 90°
Notice how all the peak-value formulas look like good old “V = IR” → the reactances XL and XC
describe the effective resistance of inductors and capacitors in AC circuits.
We now set the circuit into oscillation by “stimulating” it with a brief pulse from some external
source of EMF. The result is that the peak voltage across the capacitor is VC,max = 120 V.
(f) What is the peak current Imax?
1
= 11.5 kHz
LC
I
From part (c) VC,max = max → I max = ωC VC,max
ωC
= (11.5 × 103 Hz )( 0.025 × 10−6 F ) (120 V ) = 34.5 mA
From part (a) ω =
L = 300 mH
C = 0.025 µF
VC,max = 120 V
L
C
(g) What are the maximum energies UL,max and UC,max stored in the inductor and capacitor
respectively?
UL,max = 1/2 L Imax2 = 0.5*(0.3)*(0.03452) = 0.18 mJ
UC,max = 1/2 C (VC,max)2 = 0.5*(0.025×10-6)*(1202) = 0.18 mJ
3
(h) Determine time-dependent expressions UL(t) and UC(t) for the two stored energies, and
add them together to find the total stored energy U(t). Finally, plot your results for all
three functions.
Something like:
t
π/2ω
π/ω
3π/2ω
2π/ω
Utotal(t) = 0.18 mJ, UC is a sin2(ωt) function while UL is a cos2(ωt) function
(i) How are UL,max, UC,max, and Umax related to each other?
Knowing this relation is extremely helpful in solving LC circuit problems!
This is no accident but always happens in RC circuits
2
All are equal
U C ,max
2
I max
1
1
1 ⎛ I max ⎞
2
⎛1⎞
2
= C VC ,max = C ( X C I max ) = C ⎜
⎟ =
⎜ ⎟
2
2
2 ⎝ ωC ⎠
2C ⎝ ω ⎠
2
2
2
2
I2 ⎛ 1 ⎞
I max
LI max
= max ⎜
=
=
= U L ,max
LC
2C ⎝ 1 / LC ⎟⎠
2C
2
4