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

Answer:

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  

$V = V_{o} (1-e^{\frac{-t}{T}})$

$4 = 13 (1-e^{\frac{-3}{T}})$

T = 8.2 sec  

Time constant is given as  

T = RC  

8.2 = (9 x 10⁻⁶) R  

R = 9.1 x 10⁵ ohm