Question

2V
V
(3)
(4)
R
R
A charged capacitor is connected with a resistor.
After how many time constants, does the energy
of capacitor becomes 1/10 th of its initial value ?
TOP BOAH-A\TARGET\PHYLENG MODULE_043
(1) 2.3
(3) 0.69
(2) 1:15
(4) None​

Answer 1

Answer:

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Explanation:

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Answer 2

V=Q^2/C

Q=√C=Q0/√10........equation(i)

Q=Q0e^-t/c........ equation (ii)

Put equation (i) and (ii) equal to e

ach other.

Q0/√1=Q0e^-t/c

Now Q0 and Q0 will cancel out.

So, 1/√10=1/e^t/c

√10=e^t/c

Multiply by 'ln' on both sides.

ln√10=ln e^t/c

1/2ln10=t/c

11/2×2.303= t/c

t=1.55c

@krishhhhhhna