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(D) In the steady state,the capacitors act as open circuits. The current $i$ flows only through the middle resistor and the battery circuit.The equiva

Vedclass · Accepted Answer

$\frac{ECR}{2(R + r)}$

Vedclass · Answer

(D) In the steady state,the capacitors act as open circuits. The current $i$ flows only through the middle resistor and the battery circuit.The equivalent resistance of the circuit is $R_{eq} = R + r$.The current flowing through the middle branch is $i = \frac{E}{R + r}$.The potential difference across the middle branch (points $a$ and $b$) is $V_{ab} = iR = \frac{ER}{R + r}$.The top branch consists of two capacitors $C$ in series,which is equivalent to a single capacitor of capacitance $C_{eq} = \frac{C}{2}$.Since the top branch is connected in parallel to the middle branch,the potential difference across the combination of two capacitors is $V_{ab} = \frac{ER}{R + r}$.The charge $q$ on each capacitor in the series combination is the same as the charge on the equivalent capacitor:$q = C_{eq} V_{ab} = \left( \frac{C}{2} \right) \left( \frac{ER}{R + r} \right) = \frac{ECR}{2(R + r)}$.

Four identical capacitors and three identical resistors are connected to a battery with $emf$ $E$ and internal resistance $r$ as shown in the figure. Find the charge on each capacitor.

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