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An inductance `L` and a resistance `R` are first connected to a battery. After some time the battery is disconnected but `L` and `R` remain connected in a closed circuit. Then the current reduces to `37%` of its initial value inA. `R//L`B. `R//L`C. `L//R`D. `1//LR` |
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Answer» Correct Answer - C `i_(0)=E//R` `i=i_(1)e^(-(Rt)/(L))` `0.37i_(0)=i_(0)e^(-(Rt)/(L))` `e^(-(Rt)/(L))=(1)/(0.37)=2.7` `(Rt)/(L)=1n(2.7) implies (Rt)/(L)=1 implies t=(L)/(R)` |
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