Question

An inductor and two capacitors are connected in the circuit as shown in fig. Initially capacitor A has no charge and capacitor B has CV charge. Assume that the circuit has no resistance at all. At $t=0$, switch S is closed, then [given $LC=\frac{2}{{\pi }^2\times {10}^4}{s}^2$ and $CV=100mC$]

• A. when current in the circuit is maximum, charge on each capacitor is same
• B. when current in the circuit is maximum, charge on capacitor A is twice the charge on capacitor B
• C. $q=50(1+cos100\pi t)mC$, where q is the charge on capacitor B at time t
• D. $q=50(1-cos100\pi t)mC$, where q is the charge on capacitor B at time t

Answer 1

Answer: (A), (C)

when current in the circuit is maximum, charge on each capacitor is same
$q=50(1+cos100\pi t)mC$, where q is the charge on capacitor B at time t

't' seconds after the circuit is closed, its state will be as shown in the figure. Here, 'q' is dependent on time and 'i' is the current in the circuit.

$i=\frac{dq}{dt}$

Applying Kirchoff's Law in a closed loop,

$-\frac{q}{C}-L\frac{di}{dt}+\frac{CV-q}{C}=0$

Simplifying, we get

$\frac{V}{L}=\frac{2q}{LC}+\frac{d^{2}q}{d{t}^2}$

The solution for this second order differential equation can be written as

$q=Asin(\omega t+\varphi )+k$

$i=A\omega cos(\omega t+\varphi )$

where $\omega =(\frac{2}{LC}{)}^{\frac{1}{2}}$ and 'A','k' and $\varphi$ are constants.

Initially $q=0,i=0$

Therefore $Asin\varphi +k=0$ and $Acos\varphi =0$

Solving, we get $q=k(1-cos\omega t)$

When current is maximum, $\frac{di}{dt}=\frac{d^{2}q}{d{t}^2}=0$

Implies $k{\omega }^{2}cos\omega t=0$ i.e. $cos\omega t=0$

$q=k$

From the original equation, we get $q=\frac{CV}{2}$

Therefore, $k=\frac{CV}{2}=50mC$

Charge on both the capacitors turns out to be 50mC when the current is maximum.

Also, charge on the capacitor B is

$CV-q=50(1+cos\omega t)mC$