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The circuit shown in Fig. is switched on at t = 0. Calculate the time at which current in `R_(2)` becomes half the steady value of current. Also, calculate the energy stored in the inductor at that time. |
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Answer» As is known, final steady current in the circuit does not depend on L. `:. I_(0) = (E)/(R_(1) + R_(2)) = (3)/(2 + 3) = 0.6 A` The growth of current in RL circuit is given by Helmholtz equation `I = I_(0) [1 - e^((-R//L) t)]` For `I = I_(0)//2`, we have to find t `:. (I_(0))/(2) = I_(0) [1 - e^((-R//L)t)]` `(1)/(2) = 1 - e^(-Rt//L)` `e^(-Rt//L) = 1 - (1)/(2) = (1)/(2)` `- (Rt)/(L) = log_(e) 1 - log_(e) 2 = 0 - 2.3026 xx 0.3010` `= - 0.6931` `t = 0.931 xx (L)/(R ) = (0.931 xx 10 xx 10^(-3))/(2 + 3)` `t = 1.386 xx 10^(-3) s` Now, `I = I_(0)//2 = 0.6//2 = 0.3 A` `:.` Energy store in the inductor `U = (1)/(2) LI^(2) = (12)/(2) (10 xx 10^(-3)) (0.3)^(2)` |
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