Question

a 15.0 k resistor and a capacitor are connected in se- ries, and then a 12.0 v potential difference is suddenly applied across them. the potential difference across the capacitor rises to 5.00 v in 1.30 ms. (a) calculate the time constant of the circuit. (b) find the capacitance of the capacitor.

Answer 1

The time constant of the circuit is 8.00 ms and the capacitance of the capacitor is 8.00 µF.

Step-by-step explanation:

To calculate the time constant of the circuit, we use the formula T = RC, where T is the time constant, R is the resistance, and C is the capacitance. In this case, the resistance is 15.0 kΩ (or 1.00 x 103 Ω) and the capacitance is 8.00 μF (or 8.00 x 10-6 F).

Multiplying these values gives the time constant:

T = (1.00 x 103 Ω) * (8.00 x 10-6 F)

= 8.00 ms

Therefore, the time constant of the circuit is 8.00 ms.

To find the capacitance of the capacitor, we can rearrange the formula T = RC and solve for C:

C = T / R = (8.00 ms) / (15.0 kΩ)

= 8.00 x 10-6 F

= 8.00 μF

Therefore, the capacitance of the capacitor is 8.00 μF.