Five capacitors of capacitances C_1 = C_2 = C | vedclass

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(D) The circuit is a balanced Wheatstone bridge because $\frac{C_1}{C_2} = \frac{C_4}{C_3} = \frac{10}{10} = 1$.Since the bridge is balanced,no charge

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$5 \mu F$,$250 \mu C$ on $C_1, C_2, C_3, C_4$ and $0 \mu C$ on $C_5$

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(D) The circuit is a balanced Wheatstone bridge because $\frac{C_1}{C_2} = \frac{C_4}{C_3} = \frac{10}{10} = 1$.Since the bridge is balanced,no charge flows through the central capacitor $C_5$.Therefore,the circuit simplifies to two parallel branches,each containing two capacitors in series.The left branch consists of $C_1$ and $C_2$ in series,and the right branch consists of $C_4$ and $C_3$ in series.Equivalent capacitance of each branch: $C_{	ext{branch}} = \frac{10 	imes 10}{10 + 10} = 5 \mu F$.Total equivalent capacitance: $C_{	ext{eq}} = 5 \mu F + 5 \mu F = 10 \mu F$.Total charge supplied by the battery: $Q = C_{	ext{eq}}V = 10 \mu F 	imes 50 \ V = 500 \mu C$.Since the branches are identical,the charge divides equally: $250 \mu C$ flows through each branch.Thus,$250 \mu C$ is on each capacitor $C_1, C_2, C_3, C_4$ and $0 \mu C$ is on $C_5$.

Five capacitors of capacitances $C_1 = C_2 = C_3 = C_4 = 10 \mu F$ and $C_5 = 2.5 \mu F$ are connected as shown,along with a battery of $50 \ V$. Find the equivalent capacitance and the charge on the capacitors.

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