JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2020
- Hard
- Numerical
An ideal cell of emf \(10\, V\) is connected in circuit shown in figure. Each resistance is \(2\, \Omega .\) The potential difference (in \(V\)) across the capacitor when it is fully charged is

Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(8\)
Step-by-step Solution
Detailed explanation
\(R _{1}\) to \(R _{5} \rightarrow\) each \(2 \Omega\)
Cap. is fully charged
So no current is there in branch \(ADB\)
Effective circuit of current flow
\(R_{e q}=\left(\frac{4 \times 2}{4+2}\right)+2\)
\(R _{ eq }=\frac{4}{3}+2=\frac{10}{3} \Omega\)
\(i=\frac{10}{10 / 3}=3 A\)
So potential different across \(AEB\)
\(\Rightarrow 2 \times 1+2 \times 3=8 V\)
Hence potential difference across
Capacitor \(=\Delta V = V _{ AEB }=8 V\)
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