Question

What value resistor will discharge a 3.00 μF capacitor to 10.0% of its initial charge in 2.20 ms ?

Answer 1

The value of resistor will be given as :

Using an equation,   Q = Q_o e^–t/RC                      { eq.1 }      

(Q / Q₀) = e^–t/RC     

log (Q / Q₀) = - t / R C                                                                         { eq.2 }

where, t = time taken = 2.2 x 10^-3 sec

C = capacitance = 3 x 10^-6 F

inserting the values in above eq.

log (0.1) = - (2.2 x 10^-3 sec) / R (3 x 10^-6 F)

- (2.302) = - (2.2 x 10^-3 sec) / R (3 x 10^-6 F)

R = (2.2 x 10^-3 sec) / (3 x 10^-6 F) (2.302)

R = (2.2 x 10^-3 sec) / (6.906 x 10^-6 F)

R = 0.3185 x 10³

R = 318.5