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 5VBThe potential difference between terminals of 15V battery is 9VCThe potential difference V_a - V_b = - 4VDAll of the above

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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 / Ph-28 Physics paper FST 16 Jan.docx