Four capacitors with capacitances C_1 = 1 mu | vedclass

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(D) The circuit consists of two parallel branches connected across a $30 \ V$ source.Branch $1$ contains $C_1$ and $C_2$ in series. The equivalent cap

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$13$

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(D) The circuit consists of two parallel branches connected across a $30 \ V$ source.Branch $1$ contains $C_1$ and $C_2$ in series. The equivalent capacitance is $C_{eq1} = \frac{C_1 C_2}{C_1 + C_2} = \frac{1 	imes 1.5}{1 + 1.5} = \frac{1.5}{2.5} = 0.6 \ \mu F$.The charge on this branch is $q = C_{eq1} 	imes V = 0.6 \ \mu F 	imes 30 \ V = 18 \ \mu C$.The potential at point $a$ relative to $A$ is $V_A - V_a = \frac{q}{C_1} = \frac{18 \ \mu C}{1 \ \mu F} = 18 \ V$.Branch $2$ contains $C_3$ and $C_4$ in series. The equivalent capacitance is $C_{eq2} = \frac{C_3 C_4}{C_3 + C_4} = \frac{2.5 	imes 0.5}{2.5 + 0.5} = \frac{1.25}{3} = \frac{5}{12} \ \mu F$.The charge on this branch is $q' = C_{eq2} 	imes V = \frac{5}{12} \ \mu F 	imes 30 \ V = 12.5 \ \mu C$.The potential at point $b$ relative to $A$ is $V_A - V_b = \frac{q'}{C_3} = \frac{12.5 \ \mu C}{2.5 \ \mu F} = 5 \ V$.The potential difference between $a$ and $b$ is $V_a - V_b = (V_A - V_b) - (V_A - V_a) = 5 \ V - 18 \ V = -13 \ V$.The magnitude of the potential difference is $|V_a - V_b| = 13 \ V$.

Four capacitors with capacitances $C_1 = 1 \ \mu F, C_2 = 1.5 \ \mu F, C_3 = 2.5 \ \mu F$ and $C_4 = 0.5 \ \mu F$ are connected as shown and are connected to a $30 \ V$ source. The potential difference between points $a$ and $b$ is: (in $V$)

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