Question

A 1.0 μF capacitor is connected to two 1.0 kΩ resistors. All items are connected in series with a switch. Initially the switch is open and the potential difference across the capacitor is 10.0V. After the switch is closed, how long will it take for the potential difference across the capacitor to decrease to 5.0 V?

Answer 1

It will take approximately 0.693 milliseconds for the potential difference across the capacitor to decrease to 5.0V after the switch is closed.

Step-by-step explanation:

To find the time it takes for the potential difference across the capacitor to decrease to 5.0V, we can use the formula for the charging and discharging of a capacitor, which is given by the equation V = V0 * e-t/RC. In this case, the initial potential difference across the capacitor (V0) is 10.0V, and the final potential difference (V) is 5.0V. We can rearrange the formula to solve for time (t):

t = -RC * ln(V/V0)

Plugging in the values:

t = -(1.0kΩ * 1.0μF) * ln(5.0V/10.0V)

t ≈ 0.693 * 1.0ms

Therefore, it will take approximately 0.693 milliseconds for the potential difference across the capacitor to decrease to 5.0V after the switch is closed.