91Ó°ÊÓ

Ohm’s law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant

V = IR

An inductor act as a wire of infinite resistance right after the circuit is closed while it acts as a normal conducting wire after a long time of circuit being closed

Energy stored by the inductor is given by

$E=\frac{1}{2}L{I}^2$

Where L is the inductance of the inductor while I is the current through the system

Power dissipated by the system is given by

$P_L=i^2{R}_L$

Insert the given data into excel sheet and then plot the graph between$R_{ext}$ on horizontal axis and$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$t$}\right.$ on the vertical axis.

The Output is

The equation of the curve is

$\frac{1}{t}=\frac{1}{L\ln 2}\left(R_{ext}+R_L\right)$

The slope of the curve is$\frac{1}{L\ln 2}$ and $R_L\left(\frac{1}{L\ln 2}\right)$ represents the intercept with the dashed red line.

The slope of the curve is given by

$S=\frac{2.20s^{−1}−1.86s^{−1}}{8\mathrm{Î\copyright }−6\mathrm{Î\copyright }}=0.17$

Get the self-inductance of the coil by using this value of the slope

$\begin{array}{r}\frac{1}{L\ln 2}=0.17\\ L=8.48H\end{array}$

From the curve we see that the intercept is at$1.1s^{-1}$

Get the resistance using this value of intercept:

$\begin{array}{r}{R}_L\left(\frac{1}{L\ln 2}\right)=1.1\\ R_L=6.46\mathrm{Î\copyright }\end{array}$

Hence the inductance of the coil and resistance of the coil are respectively

The energy stored in the coil depends on the inductance and the current through the coil and it is given by

$U_L=\frac{1}{2}L{i}^2$

Plug in the values we get

$\begin{array}{r}{U}_L=\frac{1}{2}\left(8.48H\right)(20A{)}^2\\ U_L=1696J\end{array}$

Power dissipated is given by

$\begin{array}{r}{P}_L=i^2{R}_L\\ P_L=\left(20A{)}^2\right(6.46\mathrm{Î\copyright })\\ P_L=2584W\end{array}$

Hence energy stored is 1696J and the energy dissipated is 2584W