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

A capacitor acts as a normal conducting wire when the circuit is just closed while it acts a wire of infinite resistance 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.

Use ohm’s law to get the initial current in the circuit

$\begin{array}{r}I=\frac{V_R}{R}\\ I=\frac{25V}{50\mathrm{Î\copyright }}\\ I=0.5A\end{array}$

Using krichhoff’s rules , we het the resistance of the coil

$\begin{array}{r}iR+i{R}_L=\mathrm{Î\mu }\\ R_L=\frac{\mathrm{Î\mu }−iR}{i}\end{array}$

Now,plug the values and get

$\begin{array}{r}{R}_L=\frac{\mathrm{Î\mu }−iR}{i}\\ R_L=\frac{25\mathrm{V}−0.5\mathrm{A}\left(50\mathrm{Î\copyright }\right)}{0.5}\\ R_L=0\end{array}$

Using Kirchhoff’s laws

$\begin{array}{r}\mathrm{Î\mu }−iR−V_L=0\\ \mathrm{Î\mu }−iR−L\frac{di}{dt}=0\\ \frac{di}{i−\frac{\mathrm{Î\mu }}{R}}=−\frac{R}{L}dt\end{array}$

Integrate both sides

$\ln \left[\frac{i−\frac{\mathrm{Î\mu }}{R}}{\frac{\mathrm{Î\mu }}{R}}\right]=−\frac{R}{L}t$

Rearrange and use$V_R=iR$ to get

$V_R=\mathrm{Î\mu }\left(1−e^{−\frac{R}{L}t}\right)$

Where L/R is called the time constant $\mathrm{Ï„}$.

When time of$\mathrm{Ï„}$ has passed we get $\mathrm{Î\mu }=24V$

Plugging this value in the equation we get

$\begin{array}{r}{V}_R=\left(24V\right)\left(1−e^{−1}\right)\\ V_R=15.2V\end{array}$

From the graph this value is obtained at a time 7.5ms hence this is the time constant

$\begin{array}{r}\frac{L}{R}=7.5\mathrm{ms}\\ L=\left(7.5\mathrm{ms}\right)\left(R\right)\\ L=375\mathrm{mH}\end{array}$

The energy stored in the coil is given by

$\begin{array}{r}U=\frac{1}{2}L{i}^2\\ U=\frac{1}{2}L{\left(\frac{\mathrm{Î\mu }}{R}\right)}^2\end{array}$

Plug the values to get

$U=46.8×{10}^{-3}J$