Chapter 28, Problem 63P

(a)

To determine

The initial rate of increase of current.

Answer to Problem 63P

The initial rate of increase of current is $\text{3}\text{kA}/\text{s}$ .

Explanation of Solution

Given:

The inductance of a coil is $L=\text{4}\text{mH}$

The resistance of a resistor is $R=150\Omega$

The voltage of ideal battery is ${\epsilon }_o=1\text{2}\text{V}$

Formula used:

The expression for the current in the circuitis given by,

  $I=\frac{{\epsilon }_o\left(1-e^{\frac{-t}{\tau }}\right)}{R}$

Calculation:

The initial rate of increase of current is calculated by differentiating the current with time.

  $\begin{array}{c}\frac{dI}{dt}=\frac{{\epsilon }_o}{R}\frac{d}{dt}\left(1-e^{\frac{-t}{\tau }}\right)\\ =\frac{{\epsilon }_o}{R}\left(-e^{\frac{-t}{\tau }}\right)\left(\frac{-1}{\tau }\right)\\ =\frac{{\epsilon }_o}{R\tau }{e}^{\frac{-t}{\tau }}\end{array}$

Solve further as,

  $\begin{array}{c}\frac{dI}{d{t}_{t=0}}=\frac{{\epsilon }_o}{R\tau }{e}^{\frac{-0}{\tau }}\\ =\frac{\text{12}\text{V}}{\text{4}\text{mH}}\\ =\text{3}\text{kA}/\text{s}\end{array}$

Conclusion:

Therefore, the initial rate of increase of current is $\text{3}\text{kA}/\text{s}$ .

(b)

To determine

The rate of increase of current when the current is half of its steady-state value

Answer to Problem 63P

The rate of increase of current when the current is half of its steady-state value is $\text{1}\text{.5}\text{kA}/\text{s}$ .

Explanation of Solution

Formula used:

The expression for the current in the circuit is given by,

  $I=\frac{{\epsilon }_o\left(1-e^{\frac{-t}{\tau }}\right)}{R}$

Calculation:

According to the given condition,

  $\begin{array}{c}I=0.5I_f\\ \frac{I}{I_f}=1-e^{\frac{-t}{\tau }}\\ 0.5-1=-e^{\frac{-t}{\tau }}\\ e^{\frac{-t}{\tau }}=0.5\end{array}$

The rate of increase of current at $I=0.5I_f$ is calculated by differentiating the current with time.

  $\begin{array}{c}\frac{dI}{dt}=\frac{{\epsilon }_o}{R}\frac{d}{dt}\left(1-e^{\frac{-t}{\tau }}\right)\\ =\frac{{\epsilon }_o}{R}\left(-e^{\frac{-t}{\tau }}\right)\left(\frac{-1}{\tau }\right)\\ =\frac{{\epsilon }_o}{R\tau }{e}^{\frac{-t}{\tau }}\end{array}$

Solve further as,

  $\begin{array}{c}\frac{dI}{dt}=\frac{0.5\times {\epsilon }_o}{L}\\ =\frac{\text{0}\text{.5}\times \text{12}\text{V}}{\text{4}\text{mH}}\\ =\text{1}\text{.5}\text{kA}/\text{s}\end{array}$

Conclusion:

Therefore, the rate of increase of current when the current is half of its steady-state value is $\text{1}\text{.5}\text{kA}/\text{s}$ .

(c)

To determine

The steady state value of current.

Answer to Problem 63P

The steady state value of current is $\text{0}\text{.08}\text{A}$ .

Explanation of Solution

Formula used:

The expression for the steady state value of current is given by,

  $I_f=\frac{{\epsilon }_o}{R}$

Calculation:

The steady state value of current is calculated as,

  $\begin{array}{c}{I}_f=\frac{{\epsilon }_o}{R}\\ =\frac{\text{12}\text{V}}{\text{150}\text{Ω}}\\ =\text{0}\text{.08}\text{A}\end{array}$

Conclusion:

Therefore, the steady state value of current is $\text{0}\text{.08}\text{A}$ .

(d)

To determine

The time taken of the current to reach $99\%$ of its steady-state value

Answer to Problem 63P

The time taken of the current to reach $99\%$ of its steady-state value is $\text{0}\text{.123}\text{ms}$ .

Explanation of Solution

Formula used:

The expression for the current in the circuit is given by,

  $I=\frac{{\epsilon }_o\left(1-e^{\frac{-t}{\tau }}\right)}{R}$

Calculation:

According to the given condition,

  $\begin{array}{c}I=0.99I_f\\ \frac{I}{I_f}=1-e^{\frac{-t}{\tau }}\\ 0.99-1=-e^{\frac{-t}{\tau }}\\ e^{\frac{-t}{\tau }}=0.01\end{array}$

Solve further as,

  $\begin{array}{c}t=-\tau \ln \left(0.01\right)\\ =-\frac{L}{R}\ln \left(0.01\right)\\ =-\frac{\text{4}\text{mH}}{\text{150}\text{Ω}}\ln \left(0.01\right)\\ =\text{0}\text{.123}\text{ms}\end{array}$

Conclusion:

Therefore, the time taken of the current to reach $99\%$ of its steady-state value is $\text{0}\text{.123}\text{ms}$ .