Nirmal Singh.
Last Activity: 11 Years ago
As in case of discharging of a capacitor though a
resistance
q = q0*e^-t/CR
i = -dq/dt = (q0/CR)*e^-t/CR
at t = 0, i = q0 / CR
we know q0 = CV
so i = CV / CR
i = V / R
i.e. the
current at t = 0 is independent of capacitance in the circuit and as here V and
R is same for both the circuit, the current at t = 0 will be same [t V/R] and maximum in both the circuits.
In case of discharging of a
capacitor through a resistance as
q = q0*e^-t/CR
t = CR loge [q0 / q]
So time taken for 50% of the charge
to leak through the resistance


i.e. the
capacitor C1 will be lose its 50% charge sooner than CO2
Hence option (b) and (d) are correct
Thanks & Regards,
Nirmal Singh
Askiitians Faculty