Question

For the circuit arrangement shown in Figure,
a. find the potential difference across each capacitor in the steady-state condition.
b. find the current through the $60\Omega$ resistor just after the instant when the key $K$ is opened.

Answer 1

In the steady state, no current passes through upper three resistors. So $I=\frac{12}{10+30}=0.3A$

Potential difference between $A$ and $B$ is $V=30\times 0.3=9V$.

Now in loop $ACDBA$:

$\frac{q}{C_1}+\frac{q}{C_2}=9$

or $\frac{q}{20}+\frac{q}{10}=9$ or $q=60\Omega$

a. Potential difference across $C_1$:

$V_1=\frac{q}{C_1}=\frac{60}{20}=3V$

Potential difference across $C_2$:

$V_2=\frac{q}{C_2}=\frac{60}{10}=6V$

b. After $K$ is opened, 12 V will be out of circuit, and capacitors will act as batteries as shown in Figure.

Now one can find $I_l=0.0375A$.