The discharging of a capacitor is given as q=q0exp[−t∕RC] RC= time constant =τ q=q0e−t∕τ If e is the capacitance of the capacitor q=CV and q=CV0 Thus, CV=CV0et∕τ V=V0e−t∕τ ......(i) From the graph (given in the problem when t=0.5,V=25 i.e., V0=25 volt. and when t = 200, V = 5 volt Thus equation (i) becomes 5=25e−200∕τ ⇒1∕5=e−200∕τ Taking loge on both sides loge