Chapter 32, Problem 32.80CP

(a)

To determine

The emf across $L$ immediately after $t=0$.

Answer to Problem 32.80CP

The emf across $L$ immediately after $t=0$ is $96\text{V}$.

Explanation of Solution

Given info: The battery emf is $18.0\text{V}$, inductance is $0.500\text{H}$, first resistance is $2.00\text{kΩ}$ and second resistance is $6.00\text{kΩ}$.

Initially switch s is closed so the total current distribute in two current one is flow in the right loop and another current flow in the left loop.

Write the expression for the total current by Kirchhoff junction rule.

$i=i_1+i_2$ (1)

Here,

$i$ is the total current.

$i_1$ is the current in the right loop.

$i_2$ is the current in the left loop.

Write the expression for the current in the right loop.

$i_1=\frac{\epsilon }{R_1}\left(1-e^{-R_{2}t/L}\right)$

Here,

$\epsilon$ is the emf of the battery.

$R_1$ is the resistance of the first resistor.

$R_2$ is the resistance of the second resistor.

$l$ is the inductance of the inductor.

Since for the steady state condition $t=0$ so,

Substitute $0$ for $t$ in equation (2).

$\begin{array}{c}{i}_1=\frac{\epsilon }{R_1}\left(1-e^{-R_2\left(0\right)/L}\right)\\ =\frac{\epsilon }{R_1}\end{array}$

Substitute $18.0\text{V}$ for $\epsilon$ and $2.00\text{kΩ}$ for $R_1$ to find $i_1$.

$\begin{array}{c}{i}_1=\frac{18.0\text{V}}{2.00\text{kΩ}\times \frac{{10}^3\Omega }{1\text{k}\Omega }}\\ =9\times {10}^{-3}\text{A}\times \frac{{10}^3\text{mA}}{1\text{A}}\\ =9\text{mA}\end{array}$

Write the expression for the current in the left loop.

$i_2=\frac{\epsilon }{R_2}$

Substitute $18.0\text{V}$ for $\epsilon$, $6.00\text{kΩ}$ for $R_2$ to find $i_2$.

$\begin{array}{c}{i}_2=\frac{18.0\text{V}}{6.00\text{kΩ}\times \frac{{10}^3\Omega }{1\text{k}\Omega }}\\ =3\times {10}^{-3}\text{A}\times \frac{{10}^3\text{mA}}{1\text{A}}\\ =3\text{mA}\end{array}$

Substitute $9\text{mA}$ for $i_1$ and $3\text{mA}$ for $i_2$ in equation (1).

$\begin{array}{c}i=9\text{mA}+3\text{mA}\\ =\text{12}\text{mA}\end{array}$

Write the expression for the voltage across the inductor by Kirchhoff loop rule after $t=0$.

$-i\left(R_1+R_2\right)+\Delta V_{\text{L}}=0$

Rearrange the term for $\Delta V_{\text{L}}$.

$\Delta V_{\text{L}}=i\left(R_1+R_2\right)$

Substitute $18.0\text{V}$ for $\epsilon$ and, $2.00\text{kΩ}$ for $R_1$, $6.00\text{kΩ}$ for $R_2$, $12\text{mA}$ for $i$ to find $\Delta V_{\text{L}}$

$\begin{array}{c}\Delta V_{\text{L}}=12\text{mA}\left(2.00\text{kΩ}+6.00\text{kΩ}\right)\\ \Delta V_{\text{L}}=96\text{V}\end{array}$

Conclusion:

Therefore, the emf across $L$ immediately after $t=0$ is $96\text{V}$.

(b)

To determine

The point of the coil that have higher potential.

Answer to Problem 32.80CP

The point $a$ is at high potential than point $b$.

Explanation of Solution

Given info: The battery emf is $18.0\text{V}$, inductance is $0.500\text{H}$, first resistance is $2.00\text{kΩ}$ and second resistance is $6.00\text{kΩ}$.

Since the current flow upwards to downward direction in the inductor coil due to some reactance of the coil the voltage drop across the inductor coil hence the point $b$  at lower potential in compare to the point $a$ of the coil.

Write the expression for the voltage drop across the inductor coil.

$\Delta V_{\text{L}}=-L\frac{di}{dt}$

Here,

$L$ is the inductance of the coil.

Conclusion:

Therefore, the point $a$ is at high potential than point $b$.

(c)

To determine

To draw: The graph of currents in $R_1$ and $R_2$ as a function of time and show values before and after $t=0$.

Answer to Problem 32.80CP

The graph of currents in $R_1$ and $R_2$ as a function of time and show values before and after $t=0$ is shown below.

Explanation of Solution

Given info: The battery emf is $18.0\text{V}$, inductance is $0.500\text{H}$, first resistance is $2.00\text{kΩ}$ and second resistance is $6.00\text{kΩ}$.

The currents in $R_1$ and $R_2$ are shown below. After $t=0$, the current in $R_1$ decreases from an initial values of $9\text{mA}$ according to $i=I_1{e}^{-Rt/L}$. Take the original current direction as positive in each resistor the current decreases from $9\text{mA}$ to zero.

In $R_2$ the currents jumps from $3.00\text{mA}$ to $-9.00\text{mA}$ upwards and then decrease in magnitude to zero. The time constant of each decay is $1/e=36.8\%$ of the original value.

Figure (I)

(d)

To determine

The time at which the value of current in $R_2$ become $2.00\text{mA}$.

Answer to Problem 32.80CP

The time at which the value of current in $R_2$ become $2.00\text{mA}$ is $75.2\text{μs}$.

Explanation of Solution

Given info: The battery emf is $18.0\text{V}$, inductance is $0.500\text{H}$, first resistance is $2.00\text{kΩ}$ and second resistance is $6.00\text{kΩ}$.

Formula to calculate the current in the circuit after $t=0$ is,

$i=i_1{e}^{-\left(\left(R_1+R_2\right)t/L\right)}$

Substitute $2.00\text{kΩ}$ for $R_1$, $6.00\text{kΩ}$ for $R_2$, $9\text{mA}$ for $i_1$, $2.00\text{mA}$ for $i$,  and $0.500\text{H}$ for $L$ to find $i$

$\begin{array}{l}2.00\text{mA}=\left(9\text{mA}\right)e^{-\left(\left(2.00\text{kΩ}+6.00\text{kΩ}\right)t/0.500\text{H}\right)}\\ e^{-\left(\left(8.00\text{kΩ}\right)t/0.500\text{H}\right)}=\frac{2.00\text{mA}}{9\text{mA}}\end{array}$

Take log and solve the equation further,

$\begin{array}{c}t=7.52\times {10}^{-5}\text{s}\times \frac{{10}^{-6}\text{μs}}{1\text{s}}\\ =75.2\text{μs}\end{array}$

Conclusion:

Therefore, the time at which the value of current in $R_2$ become $2.00\text{mA}$ is $75.2\text{μs}$.