The current growth in two L-R circuits (b) an | vedclass

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(D) The expression for current growth in an $L-R$ circuit is given by $i = \frac{E}{R}(1 - e^{-t / 	au}) = I_0(1 - e^{-t / 	au})$,where $	au = \fra

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Both $(B)$ and $(C)$

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(D) The expression for current growth in an $L-R$ circuit is given by $i = \frac{E}{R}(1 - e^{-t / 	au}) = I_0(1 - e^{-t / 	au})$,where $	au = \frac{L}{R}$ is the time constant.From the graph,both curves reach the same steady-state current $I_0 = \frac{E}{R}$. Since the battery voltage $E$ is the same for both circuits,we have $I_{0,1} = I_{0,2} \Rightarrow \frac{E}{R_1} = \frac{E}{R_2}$,which implies $R_1 = R_2$.The time constant $	au$ determines the rate of growth of current. $A$ larger time constant means the current takes longer to reach its steady state. From the graph,curve $(c)$ grows more slowly than curve $(b)$,so $	au_c > 	au_b$. Since $	au = \frac{L}{R}$ and $R_1 = R_2$,we have $\frac{L_2}{R_2} > \frac{L_1}{R_1} \Rightarrow L_2 > L_1$,or $L_1 Thus,both statements $(B)$ and $(C)$ are correct.

The current growth in two $L-R$ circuits $(b)$ and $(c)$ is shown in figure $(a)$. Let $L_1, L_2, R_1$ and $R_2$ be the corresponding values in the two circuits. Then:

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