Problem: Consider the AC circuit shown in figure with parallel LC
connection, where L1 = L2 = L and C1 = C2 = C3 = C. When switch
K is closed, a sinusoidal AC power source is connected, and the current
amplitude remains constant, while the frequency is adjustable.
1. Before closing switch K, calculate the natural resonant frequency
of the circuit L1 L2 C1 C2 C3 .
2. If the charges on capacitors C1 and C2 are initially set to i10 and
i20 , respectively, find the initial relationship between i10 and i20 to
achieve resonance at a specific natural frequency.
3. Assuming that before closing switch K, the current and voltage in
the circuit are zero, calculate the voltage across capacitor C3 after
switch K is closed.
Solution:
1. Using Kirchhoff’s loop rule:
( R
1 di1 1
R
C R i 1 dt + L dt + C (i1 −Ri2 )dt = 0
1 di2 1
C (i2 − i1 )dt + L dt + C i2 dt = 0
2
L ddti21 = C1 (−2i1 + i2 )
2
L ddti22 = C1 (i1 − 2i2 )
1
Rewrite equation above in matrix form:
2i1 2 −2 1 i1
D = ω0
i2 1 −2 i2
√1 d2
Where ω0 = LC
and D = dt2 .
Equation above represents a coupled harmonic motion equation,
and its solution is:
(
i1 = i10 ejωt
i2 = i20 ejωt
Substituting and solving yields:
2
ω − 2ω02 ω02 i10
2 2 2 =0
ω0 ω − 2ω0 i20
The condition for a non-zero solution is that the determinant of
the coefficient matrix is zero, i.e.,
ω 2 − 2ω02 ω02
=0
ω02 ω 2 − 2ω02
√
Solving this equation results in ω = ω0 and ω = 3ω0 .
√
2. Substituting ω = ω0 , 3ω0 into equation in part 1 respectively, we
can get:
−ω02 ω02 i10
ω = ω0 ⇒ = 0 ⇒ i10 = i20
ω02 −ω02 i20
√
2 2
ω0 ω0 i10
ω = 3ω0 ⇒ = 0 ⇒ i10 = −i20
ω02 ω02 i20
2
3. Let s = jω, where s is a complex frequency. The capacitive reac-
1
tance is given by XC = sC , and the inductive reactance is XL = sL.
Therefore, the above circuit can be equivalently represented as fol-
lows:
In the diagram, i1 and i2 represent the currents passing through
L1 and L2 , respectively.
Let Z1 be the total equivalent impedance of the parallel circuit
with C1 reactance:
L2 C 2 s4 +3LCs2 +1
Z1 = sC(LCs2 +2)
Considering the current passing through Z1 , we have:
1
LCs2 +2
i1 = sC
1
Z1 + sC
I= (LCs2 +3)(LCs2 +1) I
Using the same method, we can find:
1
1
i2 = 1
sC
1 i1 = (LCs2 +3)(LCs2 +1) I
sC +sL+ sC
The voltage across C3 is:
1
I0 ω04 sin ωt
VC3 = Re i2 × sC = ωC (ω 2 −3ω 2 )(ω 2 −ω 2 )
0 0
