Time constant of a capacitive circuit

The correct answer is: A. Increases with the decrease of capacitance and decrease of resistance.

The time constant of a capacitive circuit is the amount of time it takes for the voltage across the capacitor to reach 63.2% of its final value. It is given by the equation $\tau = RC$, where $R$ is the resistance and $C$ is the capacitance.

As the capacitance decreases, the time constant decreases. This is because the capacitor has less charge to store, so it can charge and discharge more quickly. As the resistance increases, the time constant also increases. This is because the resistor provides more opposition to the flow of current, so it takes longer for the capacitor to charge and discharge.

Therefore, the time constant of a capacitive circuit increases with the decrease of capacitance and decrease of resistance.

Here is a more detailed explanation of each option:

• Option A: Increases with the decrease of capacitance and decrease of resistance. This is the correct answer, as explained above.
• Option B: Increases with the decrease of capacitance and increase of resistance. This is incorrect, as the time constant decreases with increasing resistance.
• Option C: Increases with the increase of capacitance and decrease of resistance. This is incorrect, as the time constant decreases with decreasing capacitance.
• Option D: Increase with increase of capacitance and increase of resistance. This is incorrect, as the time constant increases with decreasing capacitance and decreasing resistance.