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An inductor of 3H is connected to a battery of emf 6V through a resistance of 100 Omega. Calculate the time constant. What will be the maximum value of current in the circuit ? |
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Answer» SOLUTION :Given that L = 3H.E=6V.R=100 `OMEGA` Time constant `tau_(L)=(L)/(R)=(3)/(100)=0.03` sec Maximum Current `I_(0)=(E)/(R)=(6)/(100) "AMP"=0.06`amp |
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