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JEE Mains · Physics · STD 12 - Current electricity
  • JEE Mains 2017
  • Medium
  • MCQ

In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance \(C\) will be

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Answer & Solution

Correct Answer

(C) \(CE\)\(\;\frac{{\;{r_2}}}{{\left( {r + {r_2}} \right)}}\)

Step-by-step Solution

Detailed explanation

In steady state, flow of current through capacitor will be zero.

Current through the circuit,

\(i=\frac{E}{r+r_{2}}\)

Potential difference through capacitor

\(V_{c}=\frac{Q}{C}=E-i r=E-\left(\frac{E}{r+r_{2}}\right) r\)

\(\therefore \quad Q=C E \frac{r_{2}}{r+r_{2}}\)

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