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an inductor of 3h is connected to abattery of emf 6v through a resistance of 100calculate the time constant what will be the maxmimum value of current in the circuit.

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Give that L=3H,E=6V,R=100Ω
Time constant \tau _{L} = \frac{L}{R}= \frac{3}{100}= 0.03 \ sec
Maximum Current I_0= \frac{E}{R} = \frac{6}{100} = 0.06 \ amp

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Satyajeet Kumar

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