Question

In the Fig 20.46 shown the potential drop across $3\mu F$ capacitor when switch S is open and switch s is closed is :

• A. 9 V, 8 V
• B. 9 V, 9 V
• C. 6 V, 8 V
• D. 12 V, 8 V

Answer 1

Answer: (D)

12 V, 8 V

When switch S is open, $18V$ will drop across $C_1=3\mu F$ and $C_2=6\mu F$.

Thus, potential across $3\mu F$ is $V_1=\frac{V{C}_2}{C_1+C_2}=\frac{18\times 6}{3+6}=12V$.

When the switch S is closed potential drop across $8\Omega$ resistor is the potential drop across $3\mu F$.

The current through each resistor is $I=\frac{18}{8+10}=1A$.

Thus voltage across $8\Omega$ resistor $V^{\prime }=IR=1\times 8=8V$

Therefore , the potential drop across $3\mu F=8V$.