Question

A 7.8 µF capacitor is charged by a 9.00 V battery through a resistance R. The capacitor reaches a potential difference of 4.20 V at a time 3.21 s after charging begins. Find R.

Answer 1

Answer:

655128 ohm

Explanation:

C = Capacitance of the capacitor = 7.8 x 10⁻⁶ F  

V₀ = Voltage of the battery = 9 Volts  

V = Potential difference across the battery after time "t" = 4.20 Volts  

t = time interval = 3.21 sec  

T = Time constant

R = resistance  

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

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

$4.20 = 9 (1-e^{\frac{-3.21}{T}})$

T = 5.11 sec  

Time constant is given as  

T = RC  

5.11 = (7.8 x 10⁻⁶) R  

R = 655128 ohm