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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