Question

3. A capacitor of luF is first charged and then discharged through a
resistance of 1 M2. Calculate the time in which the charge on the
capacitor will fall to 36.8% of its initial value.
[Ans: 1 second]​

Answer 1

Explanation:

We know that for discharging of capacitor, where t is the time of discharge, Q=Q

o

e

(

RC

−t

)

Where Q

o

is the initial charge in the capacitor. As we know, Q=CV.

So, CV=CV

o

e

(

RC

−t

)

According to question, V=

2

V

o

.

Therefore,

2

V

o

=V

o

e

RC

−t

=>

2

1

=e

RC

−t

=>log0.5=

RC

−t

=>−0.301=

10×10

6

×0.1×10

−6

−t

=>−t=−0.301×(10×0.1)

=>t=0.301s

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