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An inductor of 2 H and a resitance of `10Omega` are conncts in series with a bttery of 5 V. The intial rate of change of current isA. `0.5As^(-1)`B. `2.0As^(-1)`C. `2.5As^(-1)`D. `0.25As^(-1)` |
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Answer» Correct Answer - C Growth of current in the circuit is `i=i_(0)(1-e^(-Rt//L))` `rArr" "(di)/(dt)=(d)/(dt)i_(0)-(d)/(dt)(i_(0)e^(-Rt//L))` `therefore" "(di)/(dt)=0-i_(0)(-(R)/(L))e^(-Rt//L)=(i_(0)R)/(L)e^(-Rt//L)` Initially, t = 0, `therefore" "(di)/(dt)=(i_(0)R)/(L)=(E)/(L)=(5)/(2)=2.5As^(-1)` |
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