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Show that the sum of instantaneous currents during growth and decay of current in LR circuit is independent of time. |
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Answer» During growth of current in LR circuit, `I_(g) = I_(0) (1 - e^(-Rt//L))` and during decay of current in LR circuit, `I_(d) = I_(0) e^(-Rt//L)` Sum of the currents at any instant t, `I_(g) + I_(d) = I_(0) (1 - e^(-Rt//L) + e^(-Rt//L)) = I_(0)` which is constant, and is independent to time. |
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