Question

To determine the inductance of an unmarked inductor, you set up the circuit shown in FIGURE. After moving the switch from$a$to$b$at $t=0s$, you monitor the resistor voltage with an oscilloscope. Your data are shown in the table:

Use an appropriate graph of the data to determine the inductance.

Short answer

Inductance,$L=40\mu \mathrm{H}$

Step-by-step solution

Introduction

Find an equation that relates the voltage across the resistor $\Delta V_R$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

$I_o=\frac{\epsilon }{R}=\frac{9\mathrm{V}}{100\Omega }=0.09\mathrm{A}$

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

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

Explanation by Graph

Now, we can use this expression into equation to get

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

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

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

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

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

Value of Inductance

The slope of the graph is calculated by

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

Value of L to be,

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