Three capacitors C_1, C_2 and C_3 are connect | vedclass

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(D) From the circuit diagram, capacitors $C_1$ and $C_2$ are connected in parallel.Let the equivalent capacitance of the parallel combination be $C_p

Vedclass · Accepted Answer

$\frac{(C_1 + C_2)V}{(C_1 + C_2 + C_3)}$

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(D) From the circuit diagram, capacitors $C_1$ and $C_2$ are connected in parallel.Let the equivalent capacitance of the parallel combination be $C_p = C_1 + C_2$.Now, the circuit consists of $C_p$ and $C_3$ connected in series with the voltage source $V$.In a series circuit, the voltage is divided in the inverse ratio of the capacitances.The voltage across $C_3$ is given by the voltage divider rule for capacitors:$V_3 = V 	imes \frac{C_p}{C_p + C_3}$Substituting $C_p = C_1 + C_2$ into the equation:$V_3 = V 	imes \frac{C_1 + C_2}{C_1 + C_2 + C_3}$Thus, the voltage across $C_3$ is $\frac{(C_1 + C_2)V}{(C_1 + C_2 + C_3)}$.

Three capacitors $C_1$, $C_2$ and $C_3$ are connected to a voltage source $V$ as shown in the figure. The voltage across $C_3$ will be:

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