JEE Mains · Physics · STD 12 - 3. current electricity

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

A \(CE\)
B \(CE\)\(\;\frac{{\;{r_1}}}{{\left( {{r_2} + r} \right)}}\)
C \(CE\)\(\;\frac{{\;{r_2}}}{{\left( {r + {r_2}} \right)}}\)
D \(CE\)\(\;\frac{{\;{r_1}}}{{\left( {{r_1} + r} \right)}}\)

Verified Solution

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

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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