# LC circuit oscillation

126 views

Capacitor $C_1$ is connected to point $A$ at $t=0$.The. switch is flipped to position $B$ at $t_1=\frac \pi3\sqrt {LC}$.Capacitor $C_2$ was uncharged while capacitor $C_3$ was charged to potential difference $\epsilon$ in the polarity shown before flipping the switch to $B$.The question is to find out charge on $C_3$ at time $t_2=t_1+\frac{53\pi}{180} \sqrt{\frac{LC}{3}}$
All the capacitors have same capacitance $C$

In this question I know the initial charges on capacitor but couldn't get what happens after time $t_1$ and $t_2$

edited May 12, 2017
Is it $E$ or $\epsilon$ on $C_3$ ?
It is the same thing.
Are all the capacitors the same value (C)?
Sorry @sammygerbil i forgot to mention that.yes the capacitors are all identical with same capacitance $C$
@sammygerbil I tried writing Kirchoff equation before time t1.It is $€=Ldi/dt+q/c$ writing di/dt as $d^2q/dt^2$ I couldnit bring it in form of known differential equation .Have you any ideas regarding this?Thanks.

1 vote

Both loops oscillate without any loss of energy.

In loop A the voltage across $C_1$ varies sinusoidally between $0$ and $2\epsilon$. The angular frequency of oscillation is $\omega_0=\frac{1}{\sqrt{LC}}$. So the voltage across $C_1$ at time $t_1$ after connection with the battery is $\epsilon (1-\cos\omega_0 t_1)$. See transient current in an LC circuit with a DC supply.

(This is similar to a mass on a spring in a uniform gravity field, with the mass being dropped when the spring is at its relaxed length. The mass oscillates around the equilibrium position for which $kx_0=mg$, so the extension oscillates between $0$ and $2x_0$. )

When the switch reaches position B the PDs across $C_1$ and $C_3$ are opposed. The total PD in the loop is $V=\epsilon_1 - E$ or $E-\epsilon_1$, depending on which is bigger, where $\epsilon_1$ is the PD across $C_1$ when the switch is moved from A to B.

(Query: Are $E$ and $\epsilon$ different? Perhaps the examiner intends them to be the same? Then we can be sure that the initial PD across $C_3$ will be greater than that across $C_1$.)

Loop B oscillates with different natural frequency $\omega=\sqrt{\frac{3}{LC}}$ because the capacitors are in series.

You can (I think) use the principle of superposition to write the PD across each capacitor in loop B in the same way as in loop A.

Imagine the net PD to be contained in a separate battery (as in loop A) of opposite polarity to the net PD $V$, and the 3 capacitors to be initially uncharged. The net PD of the battery oscillates sinusoidally between the one inductor and the 3 capacitors, between $0$ and $-2V$. The voltage across each capacitor reaches a maximum of $-\frac23V$.

On top of this solution you superpose the initial values of the PD across each capacitor. So $C_1, C_2, C_3$ reach maximum PDs of $-\epsilon_1-\frac23V, -\frac23V, E-\frac23V$. (I have used $-\epsilon_1$ as the initial PD on $C_1$ because it had opposite polarity to $C_3$. So assuming that $|\epsilon_1| \lt |E|$ then the reverse polarity on $C_1$ will initially increase.)

answered May 8, 2017 by (28,896 points)
selected May 12, 2017 by Green
$\epsilon$ is same as $E$ sorry for the confusion. So the initial potential on $C_1$ is $\epsilon (1-\cos \pi /3)$ is $\epsilon /2$. So the net potential in loop $B$ is$\epsilon/2$ Hence according to your answer potential across $C_3$ should be $\epsilon + \epsilon/2(1-\cos \sqrt 3 \omega t)$ Putting value of t2 i am not getting the correct answer. Can you please suggest where i have committed mistake. Thanks
Oh yes I got it.the superposition technique works.