If, in Fig. $34-2, R=20 \Omega, L=0.30 \mathrm{H}$, and $\mathscr{E}=90 \mathrm{~V}$, what will be the current in the circuit $0.050 \mathrm{~s}$ after the switch is closed?
We are going to use the exponential equation for $i$ given on p. 374 .
The time constant for this circuit is $L / R=0.015 \mathrm{~s}$, and $i_{\infty}=\mathscr{/} R=4.5 \mathrm{~A}$. Then
$$
i=i_{\infty}\left(1-e^{-t / L / R}\right)=(4.5 \mathrm{~A})\left(1-e^{-3.33}\right)=(4.5 \mathrm{~A})(1-0.0357)=4.3 \mathrm{~A}
$$