The given circuit shows capacitors $C_1$ and $C_2$ connected in parallel, and their equivalent capacitance is then in series with capacitor $C_3$.

Step-by-Step Analysis:

1. Capacitors in Parallel ( $C_1$ and $C_2$ ): When two capacitors are in parallel, their equivalent capacitance $C_{parallel}$ is simply the sum: $C_{parallel} = C_1 + C_2$

2. Combination with Series Capacitance (with $C_3$): The equivalent capacitance of the entire circuit ( $C_{eq}$ ) when $C_{parallel}$ is in series with $C_3$ is: $\frac{1}{C_{eq}} = \frac{1}{C_{parallel}} + \frac{1}{C_3}$ Which simplifies to: $C_{eq} = \frac{(C_1 + C_2) \cdot C_3}{C_1 + C_2 + C_3}$

3. Switching Capacitors $C_1$ and $C_2$: Since the sum $C_1 + C_2$ remains unchanged regardless of switching their positions, the equivalent capacitance $C_{eq}$ also remains unchanged.

Answer

The correct option is:

• It remains the same.

Would you like more detailed explanations or have any other questions?

Five Relative Questions:

1. What happens to the equivalent capacitance if $C_1$ and $C_2$ are connected in series instead of parallel?
2. How does the equivalent capacitance change if all three capacitors were in series?
3. How do we calculate the charge stored on each capacitor in this configuration?
4. How does introducing another capacitor in parallel with $C_3$ affect the equivalent capacitance?
5. What are some practical applications of capacitors arranged in parallel and series?

Tip:

When capacitors are connected in parallel, the equivalent capacitance always increases. In series, it decreases compared to the smallest individual capacitance.