Capacitor C 1 of the capacitance 1 microfarad and capacitor C 2 of capacitance 2 microfarad are separately charged fully by a common battery. The two capacitors are then separately allowed to discharge through equal resistors at time t = 0.
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(b, d)
During decay of charge in RC circuit
I = I 0 e –t/RC
where 
when t = 0, I = 
Since potential difference between the plates is same initially therefore I same in both the cases at t = 0 and is equal to
Also q = q 0 e –t/RC . When q = 
then 
= q 0 e –t/RC
⇒ e +t/RC = 2.
= ln2
⇒ t = RC log e 2
⇒ t ∝ C. Therefore time taken for the first capacitor (1µF) for discharging 50% of Initial charge will be less.
, are the correct options
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