Physics - Capacitor Question with Solution | TestHub

Let the given circuit be analyzed by labeling its nodes. Let the input terminal be $X$ and the output terminal be $Y$.

Let the junction between the first and second capacitors in the middle branch be node $A$, and the junction between the second and third capacitors be node $B$.

Based on the circuit diagram, the five capacitors are connected as follows:
1. The top capacitor is connected between nodes $X$ and $B$.
2. The first middle capacitor is connected between nodes $X$ and $A$.
3. The second middle capacitor is connected between nodes $A$ and $B$.
4. The third middle capacitor is connected between nodes $B$ and $Y$.
5. The bottom capacitor is connected between nodes $A$ and $Y$.

This configuration forms a Wheatstone bridge between terminals $X$ and $Y$, with the capacitor between $A$ and $B$ acting as the central cross arm.

Checking the balance condition for the Wheatstone bridge by comparing the ratio of the capacitances in the adjacent arms:
$\frac{C_{XB}}{C_{BY}}=\frac{C}{C}=1$
$\frac{C_{XA}}{C_{AY}}=\frac{C}{C}=1$

Since the ratios are equal, the Wheatstone bridge is balanced. This implies that nodes $A$ and $B$ are at the same electrical potential. Consequently, no charge flows through the central capacitor $C_{AB}$, and it can be effectively removed from the circuit.

After removing the central capacitor, the circuit simplifies to two parallel branches:
1. The upper branch consists of $C_{XB}$ and $C_{BY}$ in series. Its equivalent capacitance is $C_{upper}=\frac{C\times C}{C+C}=\frac{C}{2}$.
2. The lower branch consists of $C_{XA}$ and $C_{AY}$ in series. Its equivalent capacitance is $C_{lower}=\frac{C\times C}{C+C}=\frac{C}{2}$.

The total effective capacitance $C_{eq}$ between $X$ and $Y$ is the sum of the capacitances of these two parallel branches:
$C_{eq}=C_{upper}+C_{lower}=\frac{C}{2}+\frac{C}{2}=C$.

Answer: $C$