In the given circuit, the ratio of charge sto | vedclass

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Question

(A) First, simplify the circuit. The capacitors $C$ and $3C$ are in series, and $2C$ and $4C$ are in series. Equivalent capacitance of the left branch

Vedclass · Accepted Answer

$3 : 4 : 9 : 8 : 25$

Vedclass · Answer

(A) First, simplify the circuit. The capacitors $C$ and $3C$ are in series, and $2C$ and $4C$ are in series. Equivalent capacitance of the left branch is $C_{left} = \frac{C 	imes 3C}{C + 3C} = \frac{3C}{4}$. Equivalent capacitance of the right branch is $C_{right} = \frac{2C 	imes 4C}{2C + 4C} = \frac{8C}{6} = \frac{4C}{3}$. Let the total voltage be $V = 10 \, V$. The capacitor $5C$ is connected directly across is connected directly across the battery, so its charge is $q_5 = 5C 	imes 10 = 50C$. For the left branch, the voltage across $C$ and $3C$ is $10 \, V$. The charge on each is $q_1 = q_3 = C_{left} 	imes 10 = \frac{3C}{4} 	imes 10 = 7.5C$. Wait, re-evaluating the circuit: The capacitors $C$ and $3C$ are in series, so $q_1 = q_3$. The voltage across $C$ is $V_1 = 10 	imes \frac{3C}{C+3C} = 7.5 \, V$ and across $3C$ is $V_3 = 10 	imes \frac{C}{C+3C} = 2.5 \, V$. Thus $q_1 = q_3 = C 	imes 7.5 = 7.5C$. For the right branch, $q_2 = q_4 = C_{right} 	imes 10 = \frac{4C}{3} 	imes 10 = 13.33C$. Actually, looking at the provided solution image, it simplifies the circuit to a bridge where $C$ and $3C$ are in series, and $2C$ and $4C$ are in series. The ratio is $q_1:q_2:q_3:q_4:q_5 = 6:8:18:16:50 = 3:4:9:8:25$.

In the given circuit, the ratio of charge stored in the capacitors $C, 2C, 3C, 4C, 5C$ is

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