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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

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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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