Question

A 9.0 µF capacitor is charged by a 13.0 V battery through a resistance R. The capacitor reaches a potential difference of 4.00 V at a time 3.00 s after charging begins. Find R.

Answer 1

9.1 x 10⁵ ohm

Explanation:

C = Capacitance of the capacitor = 9 x 10⁻⁶ F

V₀ = Voltage of the battery = 13 Volts

V = Potential difference across the battery after time "t" = 4 Volts

t = time interval = 3 sec

T = Time constant

R = resistance

Potential difference across the battery after time "t" is given as

`V = V_(o) (1-e^{(-t)/(T)})`

`4 = 13 (1-e^{(-3)/(T)})`

T = 8.2 sec

Time constant is given as

T = RC

8.2 = (9 x 10⁻⁶) R

R = 9.1 x 10⁵ ohm