Test Index

Capacitor

Section: Physics

Question : 4 of 19

Marks: +1, -0

Solution:

In given circuit capacitor does not conducts once it is charged. So in steady state we have following circuit.

⇒ Now, let

I=

current drawn from cell. Then we have above current distribution.

Using KVL in loop

A B C D E F G A

; we have;

\;-6 I-1 \;\frac{I}{2}-\;\frac{3I}{2}-I+9=0

\Rightarrow \;\; 9 I = 9 \; \text{ or } \; I = 1 A

Now, potential difference between

B

and

D

(using KVL);

\; V_B - 6 \times 1 - 1 \times \; \frac{1}{2} = V_D

\Rightarrow V_B - V_D = 6 + \; \frac{1}{2} = \; \frac{13}{2} = 6.5 V

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