Question

Find time after which

(a) charge on capacitor becomes half of maximum value .

(b) current becomes half at minimum value .

IIT advanced question . ​

Answer 1

Hii !!

Solution is in this given attachment !

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Hope this helps !!

@Rajkumar

Answer 2

Explanation:

(a)

We know,

q = cε(1 - e^-t/c)  ---- (1)

charge on capacitor becomes half of maximum value .

q = cε/2

Place it in (1), we get

cε/2 = cε(1 - e^-t/2)

=> (1/2) = (1 - e^-t/2)

=> e^-t/2 = 1/2

=> -(t/2) = log(1/2)

=> t = 2 log 2

(b)

We know,

I = ε/R[e^-t/2]

current becomes half at minimum value .

=> ε/2R = ε/R * e^-t/2

=> e^-t/2 = 2

=> t = 2^log2

Hope it helps!