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In the figure shown, $C_{1} = 11 μ F a n d C_{2} = 5 μ F$ , then at steady state :

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The potential difference across $C_{1}$ is $5 V$ .

The potential difference across $C_{2}$ is $2 V$ .

The potential difference between points a and b is $4 V$ .

The potential difference between the terminals of 15V battery is $9 V$ .

answer is A, C, D.

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Detailed Solution

current through $3 Ω$ : $i = \frac{15}{3 + 2} = 3 A$

so, potential across $3 Ω$ is 9 V.

$\frac{q}{C_{1}} + \frac{q}{C_{2}} = 9 + 7 = 16$ ; $q = 55$

$V_{1} = \frac{q_{1}}{C_{1}} = \frac{55}{11} = 5 V$

$V_{2} = \frac{q_{2}}{C_{2}} = \frac{55}{5} = 11 V$

$V_{a} - V_{b} = 11 - 7 = 4 V$

potential across $3 Ω$ is the potential across the terminal of the battery.

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