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An inductor (L=0.03H) and a resistor (R=0.15kOmega) are connected in series to a battery of 15 V emf in a circuit shown below. The key K_1 has been kept closed for a long time. Then at t=0 , K_1 is opened and key K_2 is closed simulatenously. At t=1 ms, the current in the circuit will be (e^5cong150) |
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Answer» 100 mA `I_0=e/R=15/(0.15times10^3)=0.1A` Here, t=1 ms =`10^-3` s Time CONSTANT of the LR circuit, `t_0=L/R=0.03/(0.15times10^3)=2times10^-4s` `thereforet/t_0=10^-3/(2times10^-4)=5` `thereforeI=I_0e^(-t//t_0)=0.1e^-5=0.1/150=0.67times10^-3A` =0.67 mA |
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