91Ó°ÊÓ

Find an equation that relates the voltage across the resistor $\mathrm{Δ}{V}_R$and the time $t$. At time $t=0$, the current doesn't flow through the inductor, so the maximum current in the circuit is calculated from Ohm's law by

$I_o=\frac{\mathrm{Î\mu }}{R}=\frac{9\mathrm{V}}{100\mathrm{Î\copyright }}=0.09\mathrm{A}$

The current in the inductor changes according to the equation

$I=I_o{e}^{-\frac{R}{L}t}$

$I$is the same current that flows through $R$and it is calculated by

$I=\frac{V_R}{R}$

Now, we can use this expression into equation to get

$\frac{V_R}{R}=I_o{e}^{-\frac{R}{L}t}$

$\frac{V_R}{100\mathrm{Î\copyright }}=(0.09\mathrm{A})e^{-\frac{R}{L}t}$

$V_R=9e^{-\frac{R}{L}t}$

$\ln V_R=\ln 9-\frac{1}{L}Rt$

we use the given data to draw a graph between $\ln V_R$and $t$where the slope of the graph is$\frac{1}{L}$

The slope of the graph is calculated by

$-\frac{1}{L}=$Slope $=\frac{1.65\mathrm{V}-1.40\mathrm{V}}{15×{10}^{-6}\mathrm{s}-25×{10}^{-6}\mathrm{s}}=-25000\mathrm{H}$

Value of L to be,

$L=\frac{1}{25000{\mathrm{H}}^{-1}}=40×{10}^{-6}\mathrm{H}=40\mathrm{μ}\mathrm{H}$