Given,
The resistance of the resistor, R=20 Ω
The inductance of the inductor, L=20 mH
The battery voltage, V=20 V
The time instant, t=2 ms=2×10⁻³ s
The instantaneous current through an LR circuit is given by,
[tex]I=\frac{V}{R}(1-e^{\frac{-Rt}{L}})[/tex]
On substituting the known values,
[tex]\begin{gathered} I=\frac{20}{20}(1-e^{\frac{20\times2\times10-3}{20\times10-3}}) \\ =6.39\text{ A} \end{gathered}[/tex]
Thus the instantaneous current through the circuit is 6.39 A
