Chapter 28, Problem 59P

(a)

To determine

The current and its rate of change at $t=0$ .

Answer to Problem 59P

The current and its rate of change at $t=0$ is $0\text{and}25\text{kA}/\text{s}$ respectively.

Explanation of Solution

Given:

The resistance and self-inductance of the coil is $8.00\text{Ω}\text{and}4.00\text{mH}$ respectively.

The voltage of the battery is $\text{100}\text{V}$ .

Formula used:

The expression for the current is given by,

  $I=I_f\times \left(1-e^{{}^{\frac{-t}{\tau }}}\right)$

Here,

  $I_f=\frac{{\epsilon }_o}{R}$

And.

  $\tau =\frac{L}{R}$

Here $I_f$ is the final current and $\tau$ is time constant.

Calculation:

The final current in a coil is calculated as,

  $\begin{array}{c}{I}_f=\frac{{\epsilon }_o}{R}\\ =\frac{\text{100}\text{V}}{\text{8}\text{.00}\text{Ω}}\\ \text{=12}\text{.5}\text{A}\end{array}$

The time constant is calculated as,

  $\begin{array}{c}\tau =\frac{L}{R}\\ =\frac{\text{4}\text{.00}\text{mH}}{\text{8}\text{.00}\text{Ω}}\\ =\text{0}\text{.5}\text{ms}\end{array}$

The current is calculated as,

  $\begin{array}{c}I=I_f\times \left(1-e^{{}^{\frac{-t}{\tau }}}\right)\\ I_0=\left(12.5\text{A}\right)\times \left(1-e^{{}^{\frac{-0}{0.5}}}\right)\\ =0\end{array}$

The rate of current with time is calculated as,

  $\begin{array}{c}\frac{dI}{dt}=\frac{d}{dt}\left(12.5\text{A}\right)\times \left(1-e^{{}^{\frac{-t}{0.5\text{ms}}}}\right)\\ \frac{dI}{d{t}_{t=0}}=\left(12.5\text{A}\right)\times \left(1-e^{{}^{\frac{-0}{\text{0}\text{.5}\text{ms}}}}\right)\times \frac{-1}{0.5\text{ms}}\\ =25\times {10}^3\text{A}/\text{s}\left(\frac{{10}^{-3}\text{kA}}{1\text{A}}\right)\\ =\text{25}\text{k}\text{A}/\text{s}\end{array}$

Conclusion:

Thereforethe current and its rate of change at $t=0$ is $0$ and $25\text{kA}/\text{s}$ respectively.

(b)

To determine

The current and its rate of change at $t=\text{0}\text{.100}\text{ms}$

Answer to Problem 59P

The current and its rate of change at $t=\text{0}\text{.100}\text{ms}$ is $2.27\text{A}\text{and}20.5\text{kA}/\text{s}$ .

Explanation of Solution

Formula used:

The expression for the current is given by,

  $I=I_f\times \left(1-e^{{}^{\frac{-t}{\tau }}}\right)$

Here,

  $I_f=\frac{{\epsilon }_o}{R}$

And.

  $\tau =\frac{L}{R}$

Here $I_f$ is the final current and $\tau$ is time constant.

Calculation:

The current at time $t=0.100\text{ms}$ is calculated as,

  $\begin{array}{c}I=I_f\times \left(1-e^{{}^{\frac{-t}{\tau }}}\right)\\ I_0{}_{.1}=\left(12.5\text{A}\right)\times \left(1-e^{{}^{\frac{\text{-0}\text{.1}\text{ms}}{\text{0}\text{.5}\text{ms}}}}\right)\\ =2.27\text{A}\end{array}$

The rate of the current with time is calculated as,

  $\begin{array}{c}\frac{dI}{dt}=\frac{d}{dt}\left(12.5\text{A}\right)\times \left(1-e^{{}^{\frac{-t}{0.5\text{ms}}}}\right)\\ \frac{dI}{d{t}_{t=0.1}}=\left(12.5\text{A}\right)\times \left(-e^{{}^{\frac{\text{-0}\text{.1}\text{ms}}{\text{0}\text{.5}\text{ms}}}}\right)\times \frac{-1}{0.5\text{ms}}\\ =20.5\text{kA}/\text{s}\end{array}$

Conclusion:

Therefore,the current and its rate of change at $t=0.1\text{ms}$ is $2.27\text{A}$ and $20.5\text{kA}/\text{s}$ respectively.

(c)

To determine

The current and its rate of change at $t=0.5\text{ms}$

Answer to Problem 59P

The current and its rate of change at $t=0.5\text{ms}$ is $7.9\text{A}\text{and}9.2\text{kA}/\text{s}$ respectively.

Explanation of Solution

Formula used:

The expression for the current is given by,

  $I=I_f\times \left(1-e^{{}^{\frac{-t}{\tau }}}\right)$

Here,

  $I_f=\frac{{\epsilon }_o}{R}$

And.

  $\tau =\frac{L}{R}$

Here $I_f$ is the final current and $\tau$ is time constant.

Calculation:

The current at time $t=0.500\text{ms}$ is calculated as

  $\begin{array}{c}I=I_f\times \left(1-e^{{}^{\frac{-t}{\tau }}}\right)\\ I_0{}_{.5}=\left(12.5\text{A}\right)\times \left(1-e^{{}^{\frac{\text{-0}\text{.5}\text{ms}}{\text{0}\text{.5}\text{ms}}}}\right)\\ =7.9\text{A}\end{array}$

The rate of the current with time is calculated as:

  $\begin{array}{c}\frac{dI}{dt}=\frac{d}{dt}\left(12.5\text{A}\right)\times \left(1-e^{{}^{\frac{-t}{0.5\text{ms}}}}\right)\\ \frac{dI}{d{t}_{t=0.5}}=-\left(12.5\text{A}\right)\times \left(e^{{}^{\frac{\text{-0}\text{.5}\text{ms}}{\text{0}\text{.5}\text{ms}}}}\right)\times \frac{-1}{0.5\text{}\text{ms}}\\ =9.2\text{kA}/\text{s}\end{array}$

Conclusion:

Therefore, the current and its rate of change at $t=0.5\text{ms}$ is $7.9\text{A}and9.2\text{kA}/\text{s}$ respectively.

(d)

To determine

The current and its rate of change at $t=1.0\text{ms}$

Answer to Problem 59P

The current and its rate of change at $t=1.0\text{ms}$ is $10.8\text{A}and3.38\text{kA}/\text{s}$ respectively.

Explanation of Solution

Formula used:

The expression for the current is given by,

  $I=I_f\times \left(1-e^{{}^{\frac{-t}{\tau }}}\right)$

Here,

  $I_f=\frac{{\epsilon }_o}{R}$

And.

  $\tau =\frac{L}{R}$

Here $I_f$ is the final current and $\tau$ is time constant.

Calculation:

The current at time $t=1.0\text{ms}$ is calculated as

  $\begin{array}{c}I=I_f\times \left(1-e^{{}^{\frac{-t}{\tau }}}\right)\\ I_{1.0}=\left(12.5\text{A}\right)\times \left(1-e^{{}^{\frac{\text{-1}\text{.0}\text{ms}}{\text{0}\text{.5}\text{ms}}}}\right)\\ =10.8\text{A}\end{array}$

The rate of the current with time is calculated as:

  $\begin{array}{c}\frac{dI}{dt}=\frac{d}{dt}\left(12.5\text{A}\right)\times \left(1-e^{{}^{\frac{-t}{0.5\text{ms}}}}\right)\\ \frac{dI}{d{t}_{t=1.0}}=-\left(12.5\text{A}\right)\times \left(e^{{}^{\frac{\text{-1}\text{.0}\text{ms}}{\text{0}\text{.5}\text{ms}}}}\right)\times \frac{-1}{0.5\text{ms}}\\ =3.38\text{kA}/\text{s}\end{array}$

Conclusion:

Therefore, the current and its rate of change at $t=1.0\text{ms}$ is $10.8\text{A}\text{and}3.38\text{kA}/\text{s}$ respectively.