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A solenoid of inductance 50 mH and resistance `10 Omega` is connected to a battery of 6V. Find the time elapsed before the current acquires half of its steady - state value.A. `3.5 ms`B. `2.5 ms`C. `0.693 ms`D. `2 ms` |
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Answer» Correct Answer - A Time constant of circuit= `L/R =(50 mH)/(10)=5 ms` Current at time T is given by `i=i_(0)(1-e^(-T//(tau)))` `e^(-T//(tau)) = 1/2 or T/(tau)=log_(e)2` `T=tau log_(2)2 = 5 ms xx 0.693 = 3.2 ms`. |
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