Question

In Fig. $\mathbf{30}\mathbf{.}\mathbf{11}$, suppose that $\mathit{\epsilon }\mathbf{=}\mathbf{60}\mathit{V}$, $\mathit{R}\mathbf{=}\mathbf{240}\mathit{\Omega }$, and $\mathit{L}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{16}\mathit{H}$. With switch $\mathit{S}\mathbf{2}$open, switch $\mathit{S}\mathbf{1}$is left closed until a constant current is established. Then $\mathit{S}\mathbf{1}$is closed and $\mathit{S}\mathbf{2}$opened, taking the battery out of the circuit. (a) What is the initial current in the resistor, just after $\mathit{S}\mathbf{2}$is closed and $\mathit{S}\mathbf{1}$is opened? (b) What is the current in the resistor at$\mathit{t}\mathbf{=}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathit{s}$? (c) What is the potential difference between points b and c at $\mathit{t}\mathbf{=}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathit{s}$? Which point is at a higher potential? (d) How long does it take the current to decrease to half its initial value?

Short answer

$$\begin{gathered} \mathrm{a})\mathrm{i}=0.25\mathrm{A} \\ \mathrm{b})\mathrm{i}=0.137\mathrm{A} \\ \mathrm{c})\mathrm{V}=32.9\mathrm{V} \\ \mathrm{d})\mathrm{t}=0.462\mathrm{ms} \end{gathered}$$

Step-by-step solution

Calculate the initial current

The formula to find the current in an LR circuit is $\mathit{i}\mathbf{=}{\mathit{I}}_{\mathbf{0}}{\mathit{e}}^{\mathbf{-}\frac{\mathbf{t}\mathbf{R}}{\mathbf{L}}}$.

To find the initial current, put $t=0$

localid="1664256507202" $$\begin{gathered} i=I_0{e}^{-\frac{\mathrm{tR}}{\mathrm{L}}} \\ i=I_0{e}^0 \\ i=I_0 \\ i=\frac{\epsilon }{R} \\ i=\frac{60}{240} \\ i=0.25A \end{gathered}$$

Find the current at the given time

In the formula of current, put $t=4\times {10}^{-4}s$.

$$\begin{gathered} i=I_0{e}^{-\frac{\mathrm{tR}}{\mathrm{L}}} \\ i=0.25e\frac{-4\times {10}^{-4}\times 240}{0.16} \\ i=0.137A \end{gathered}$$

Find the potential difference by multiplying current and resistance Step 3: Find the potential difference by multiplying current and resistance

$$\begin{gathered} V=IR \\ V=0.137\times 240 \\ V=32.9V \end{gathered}$$

Use the current equation to find the time for half the value

According to the situation,$i=0.5I_0$.

role="math" localid="1664256476180" $$\begin{gathered} i=0.5I_0=I_0{e}^{-\frac{tR}{L}} \\ 0.5=e^{-\frac{tR}{L}} \\ t=-\frac{L}{R}\ln 0.5 \\ t=-\frac{0.16}{240}\ln 0.5 \\ t=0.462ms \end{gathered}$$