Question

Two capacitors of capacitance , 2C and C, respectively, are connected in series with an inductor of inductance 'L' . Initially the capacitors have charge such that $V_B-V_A=4V_0$ and $V_C-V_D=V_0$.Initial current in the circuit is zero.
Find
a) The maximum current that will flow in the circuit
b) The potential difference across each capacitor at that instant,
c) The equation of the current flowing towards left in the inductor.

Answer 1

a) Applying Kirchoffs voltage loop rule in the given figure
$\frac{q_1}{2C}-\frac{q_2}{C}-L\frac{dI}{dt}=0$
or $L\frac{dI}{dt}=\left(\frac{q_1-2q_2}{2C}\right)-\frac{3q}{2C}$.............(i)

Differentiating (i) we get,
$L\frac{d^{2}I}{d{t}^2}=-\frac{3}{2C}I$.......................................(ii)

Solution of (ii) is $I=I_{0}sin(\omega t+\varphi )$
where,${\omega }^2=\frac{3}{2CL}\Rightarrow \omega =\sqrt{\frac{3}{2CL}}$

but initial current is zero , i,e at t=0,I=0, putting this above we get $I=I_0$.......................(iii)
$\Rightarrow \frac{dI}{dt}=I_0\omega cos\omega t$

Putting dI/dt in eq.(i) , $L{I}_0\omega cos\omega t=\left(\frac{q_1-2q_2}{2C}\right)-\frac{3q}{2C}$
At $t=0,q=0$; putting this in the above eq, we get
$L{I}_0\omega =\left(\frac{q_1-2q_2}{2C}\right)\Rightarrow L{I}_0\sqrt{\frac{3}{2CL}}=\left(\frac{8C{V}_0-2C{V}_0}{2C}\right)$
$\Rightarrow I_0=V_0\sqrt{\frac{6C}{L}}\to$ the maximum value of current

b) From (iii) maximum current occurs at $\omega t=\pi /2,3\pi /2,.....$etc and at these instants from (i) we get

$q=\left(\frac{q_1-2q_2}{3}\right)$
Potential difference across each capacitor at this instant
$V_1=\frac{q_1}{2C}=\frac{1}{2C}[q_1-\frac{q_1-2q_2}{3}]$
$V_1=\frac{(q_1+q_2)}{3C}=3V_0$
$V_2=\frac{q_2}{C}=\frac{1}{C}[q_2-(\frac{q_1-2q_2}{3}\left)\right]$
$V_2=\frac{(q_1+q_2)}{3C}=3V_0$

c) Equation of current
$I=I_{0}sin\omega t=V_0\sqrt{\frac{6C}{L}}sin\sqrt{\frac{3}{2CL}}t$