In this figure the resistance of the coil of galvanometer $G$ is $2\,\Omega$. The emf of the cell is $4\,V$. The ratio of potential difference across $C_1$ and $C_2$ is:
JEE MAIN 2023, Diffcult
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At steady state, current in the circuit is
$i =\frac{4 V }{6+2+8}=\frac{1}{4}\,A$
Voltage across $C_1$ is
$V _1= V _{ AC }=i(6 \Omega+2 \Omega)=\frac{1}{4} \times 8=2\,V$
Voltage across $C _2$ is
$V _2= V _{ BD }=i(2 \Omega+8 \Omega)=\frac{1}{4} \times 10=2.5\,V$
$\Rightarrow \frac{ V _1}{ V _2}=\frac{2}{2.5}=\frac{4}{5}$
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