Question bank/JEE Main/Physics

In the figure shown $C_{1} = 11\ \mu F$ and $C_{2} = 5\ \mu F$
then at steady state

Athe potential difference across C_1 is 5V.Bthe potential difference across C_2 is 2V.Cthe potential difference$\ V_{a} - V_{b} = - 4V$.Dthe potential difference between terminals of 15V battery is 9V.

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[IMAGE] at steady state
$I(3) + 1(2) = 15$
$I = 3$
kvL
$C \rightarrow D \rightarrow E \rightarrow a \rightarrow b \rightarrow C$
$V_{c} - I(3) + \frac{q}{11} - 7 + \frac{q}{5} = V_{c}$
$\frac{q}{11} + \frac{q}{5} = 7 + 3 \times 3 = 16$
$q = 55\mu c$
Now kvL for $a \rightarrow b$
$v_{a} - 7 + \frac{q}{5} = v_{b}$
$\Rightarrow V_{a} - V_{b} = 7 - \frac{q}{5}$
$= 7 - \frac{55}{5}$
$= - \frac{20}{5} = - 4V$

Source: Grass-G / P-17 Physics paper 26 Nov Second.docx