Question

An inductor with inductance L = 0.300 H and negligible resistance is connected to a battery, a switch S, and two resistors, $\mathrm{R}1=12.0\mathrm{\Omega }$and $\mathrm{R}2=16.0\mathrm{\Omega }$(Fig. P30.50). The battery has emf 96.0 V and negligible internal resistance. S is closed at t = 0. (a) What are the currents just after S is closed? (b) What are ${\mathrm{i}}_1,{\mathrm{i}}_2$and ${\mathrm{i}}_3$after S has been closed a long time? (c) What is the value of t for which ${\mathrm{i}}_3$has half of the final value that you calculated in part (b)? (d) When has half of its final value, what are ${\mathrm{i}}_1\mathrm{and}{\mathrm{i}}_2$?

Short answer

A) The current passing through${\mathrm{i}}_2={\mathrm{i}}_1=8\mathrm{A}$and${\mathrm{i}}_3=0$

b) The current flowing in${\mathrm{i}}_1=14\mathrm{A}$

C) The value of t is 13ms

D) i3 that has half of its final value is 11A

Step-by-step solution

Concept of the current passing through the resistors just after S is closed

The current passing through the resistors is given as${\mathrm{i}}_1={\mathrm{i}}_2=\frac{\mathrm{\epsilon }}{\mathrm{R}}$when${\mathrm{i}}_3=0$

${\mathrm{i}}_2={\mathrm{i}}_1=\frac{96\mathrm{V}}{12\mathrm{\Omega }}=8\mathrm{A}$

Therefore, the current passing through${\mathrm{i}}_2={\mathrm{i}}_1=$8A and${\mathrm{i}}_3=0$

Calculate i1, i2, and i3 after S has been closed a long time

When reached steady state $\left|\frac{{\mathrm{di}}_3}{\mathrm{dt}}\right|=0$,Potential across the inductor is${\mathrm{\epsilon }}_{\mathrm{L}}=-\mathrm{L}\left|\frac{{\mathrm{di}}_3}{\mathrm{dt}}\right|=0$

$$\begin{gathered} {\mathrm{i}}_2=\frac{\mathrm{\epsilon }}{{\mathrm{R}}_2}=\frac{96\mathrm{V}}{12\mathrm{\Omega }}=8\mathrm{A} \\ {\mathrm{i}}_3=\frac{\mathrm{\epsilon }}{{\mathrm{R}}_2}=\frac{96\mathrm{V}}{16\mathrm{\Omega }}=6\mathrm{A} \end{gathered}$$

By Kirchhoff’s rule,

$$\begin{gathered} {\mathrm{i}}_2={\mathrm{i}}_3={\mathrm{i}}_1 \\ 8\mathrm{A}+6\mathrm{A}=14\mathrm{A} \end{gathered}$$

Therefore, the current flowing in${\mathrm{i}}_1=14\mathrm{A}$

Calculate the value of t for which i3 has half of the final value

$$\begin{gathered} \mathrm{E}=\frac{\mathrm{L}}{{\mathrm{R}}_2}\mathrm{In}\left(\frac{-\mathrm{\epsilon }/{\mathrm{R}}_2}{{\mathrm{i}}_3-\mathrm{\epsilon }/{\mathrm{R}}_2}\right) \\ {\mathrm{i}}_3=\frac{6\mathrm{A}}{2}=3\mathrm{A} \\ \mathrm{t}=\frac{\mathrm{L}}{{\mathrm{R}}_2}\mathrm{In}\left(\frac{-\mathrm{\epsilon }/{\mathrm{R}}_2}{{\mathrm{i}}_3-\mathrm{\epsilon }/{\mathrm{R}}_2}\right) \\ =\frac{0.3}{16}\mathrm{In}\left(\frac{-96/16}{3-96/16}\right) \\ =0.0130\mathrm{s} \\ \mathrm{t}=13\mathrm{ms} \end{gathered}$$

Therefore, the value of t is $13\mathrm{ms}$

Calculate i3 that has half of its final value

${\mathrm{i}}_3$has half of its value so,${\mathrm{i}}_3=\frac{6\mathrm{A}}{2}=3\mathrm{A}$ and${\mathrm{i}}_2=\frac{\mathrm{\epsilon }}{{\mathrm{R}}_2}=\frac{96\mathrm{V}}{12\mathrm{\Omega }}=8\mathrm{A}$

By Kirchhoff’s rule,

$$\begin{gathered} {\mathrm{i}}_2={\mathrm{i}}_3={\mathrm{i}}_1 \\ 8\mathrm{A}+3\mathrm{A}=11\mathrm{A} \end{gathered}$$

Therefore, i3 that has half of its final value is 11A