JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2023
- Hard
- MCQ
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:

Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(\frac{4}{5}\)
Step-by-step Solution
Detailed explanation
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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