Question

the charge on a capacitor of capacitance 1 microfarad falls to 50 %of its initial value in 5 minutes. when the two plates of the capacitor are joined by an unknown resistance. what is the value of this resistance?

Answer 1

formula of discharging capacitor is given by, $q(t)=Q_0e^{-\frac{t}{RC}}$

where $Q_0$ is initial amount of charge on capacitor , C is capacitance of capacitor, R is resistance of resistor joined in circuit and t is time taken to discharge.

here, $q(t=5min)=\frac{Q_0}{2}$

C = 1 microfarad = 10^-6 F and t = 5 min = 300 sec

so, $\frac{Q_0}{2}=Q_0e^{-\frac{300}{10^-6R}}$

or,$\frac{1}{2}=e^{-\frac{3\times10^8}{R}}$

or, -ln2 = - 3 × 10^8/R

or, R = 3 × 10^8/ln2 ohm

R = 4.32 × 10^8 ohm