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The circuit shown in Fig. is switched on...

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.

Text Solution

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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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