Question

the charge on a capacitor becomes 1/n times of its initial value in time t,when it discharges through a circuit with constant time T the value of t is
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Answer 1

The charge on a capacitor becomes 1/n times of its initial value in time t, when it discharges through a circuit with constant time τ.

To find : The value of time t is ...

solution : let initial value of charge in the capacitor is $q_0$

at time t, charge becomes 1/n times of its initial value. i.e., $q=\frac{q_0}{n}$

using formula, $q=q_0e^{-t/\tau}$

⇒$\frac{q}{q_0}=q_0e^{-t/\tau}$

⇒$\frac{1}{n}=e^{-t/\tau}$

taking log base e both sides,

⇒$-ln(n)=-\frac{t}{\tau}$

⇒t = τln(n)

Therefore the value of time t is τln(n)

Answer 2

Answer:

During discharge the current though resistance is i=i

0

e

−t/η

where η= time constant.

Power dissipation through resistor R is P=i

2

R=i

0

2

e

−2t/η

R=P

0

e

−t/(η/2)

Thus time constant becomes

2

η

Explanation:

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