For the value of obtained in the above question, the time taken for 95% of the stored energy to be dissipated is close to

(a) 0.10 s (b) 0.15 s

(c) 0.50 s (d) 1.0 s

Step-by-Step

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E_1 is the stored energy and E_2 is the dissipated energy. So,

\frac{i_2^2}{i_1^2}=\frac{E_2}{E_1}=0.95

Therefore, i_2^2=0.95 i_1^2 \Rightarrow i_2=0.9746 i_1

When the switch is in position 2,

i(t)=i_1 e^{-(R / L) t}=2 e^{-(60 / 10) t}=2 e^{-6 t} (i)

After 95% of energy dissipation,

i(t) = 2 – 2 × 0.97 = 0.05 A (ii)

Substituting Eq. (ii) in Eq. (i), we get

0.05=2 e^{-6 t} \Rightarrow t=0.5 s

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