Chapter 29, Problem 38P

(a)

To determine

The peak current in the circuit.

Answer to Problem 38P

The required peak current is 7.94 A.

Explanation of Solution

`Given:

Inductance of inductor is $L=36\text{mH}$

Resistance of resistor is $R=40\Omega$

Ideal voltage output is $\epsilon =\left(345\text{V}\right)\times \cos \left(150\pi t\right)$

Formula used:

Expression for peak current is

  $\begin{array}{l}\epsilon -IZ=0\\ I=\frac{\epsilon }{Z}\end{array}$

  $I=\frac{\epsilon }{\sqrt{R^2+{\left(\omega L\right)}^2}}$.......... (1)

Here, $\epsilon ,R,L\text{and}\omega$ are voltage, resistance, angular frequency and inductance of inductor.

Calculation:

Substitute $\left(345\text{V}\right)\times \cos \left(150\pi t\right)$ for $\epsilon$ , $40\Omega$ for $R$ , $150\pi$ for $\omega$ , $36\text{mH}$ for $L$ in equation (1)

  $\begin{array}{l}I=\frac{345\times \cos \left(150\pi \times t\right)}{\sqrt{{\left(40\Omega \right)}^2+{\left(150\pi \times 36\times {10}^{-3}\right)}^2}}\\ I=\left(7.94\text{A}\right)\times \cos 150\pi t\end{array}$

Conclusion:

Hence, the requiredpeak current is 7.94 A.

(b)

To determine

The peak and RMS voltage across the inductor.

Answer to Problem 38P

The required peak voltage and RMS voltage are $\boxed{V_{peak}=345\text{V}}$ and $\boxed{V_{rms}=244\text{V}}$ .

Explanation of Solution

Given:

Inductance of inductor is $L=36\text{mH}$

Resistance of resistor is $R=40\Omega$

Ideal voltage output is $\epsilon =\left(345\text{V}\right)\times \cos \left(150\pi t\right)$

Formula used:

RMS current can be obtained as

  $V_{rms}=\frac{V_{peak}}{\sqrt{2}}$.......... (1)

Here, $V_{peak}$ is the peak voltage.

Calculation:

Expression for output voltage is

  $\epsilon =\left(345\text{V}\right)\times \cos \left(150\pi t\right)$

From above expression, peak voltage is found to be $345\text{V}$ i.e. $\boxed{V_{peak}=345\text{V}}$ .

Substitute $345\text{V}$ for $V_{peak}$ in equation (1)

  $\begin{array}{l}{V}_{rms}=\frac{345}{\sqrt{2}}\\ \boxed{V_{rms}=244\text{V}}\end{array}$

Conclusion:

Hence, the required peak voltage and RMS voltage are $\boxed{V_{peak}=345\text{V}}$ and $\boxed{V_{rms}=244\text{V}}$ .

(c)

To determine

The average power dissipated.

Answer to Problem 38P

  $\boxed{P_{avg}=1.3\text{kW}}$

Explanation of Solution

Given:

Inductance of inductor is $L=36\text{mH}$

Resistance of resistor is $R=40\Omega$

Ideal voltage output is $\epsilon =\left(345\text{V}\right)\times \cos \left(150\pi t\right)$

Peak current is $I_{peak}=7.94\text{A}$

Formula used:

Average power can be obtained as

  $\begin{array}{l}{P}_{avg}=I^2{}_{rms}\times R\\ P_{avg}={\left(\frac{I_{peak}}{\sqrt{2}}\right)}^2\times R\end{array}$

  $P_{avg}=\frac{1}{2}{I}^2{}_{peak}\times R$.......... (1)

Here, $I_{peak}$ is the peak current and $R$ is resistance.

Calculation:

Substitute $7.94\text{A}$ for $I_{peak}$ and $40\Omega$ for $R$ in equation (1)

  $\begin{array}{l}{P}_{avg}=\frac{1}{2}\times {\left(7.94\text{A}\right)}^2\times \left(40\Omega \right)\\ \boxed{P_{avg}=1.3\text{kW}}\end{array}$

Conclusion:

Hence, the required average power is $\boxed{P_{avg}=1.3\text{kW}}$.

(d)

To determine

The peak and average magnetic energy stored in inductor.

Answer to Problem 38P

  $\boxed{U_{peak}=1.1\text{J}}$ and $\boxed{U_{avg}=0.57\text{J}}$

Explanation of Solution

Given:

Inductance of inductor is $L=36\text{mH}$

Resistance of resistor is $R=40\Omega$

Ideal voltage output is $\epsilon =\left(345\text{V}\right)\times \cos \left(150\pi t\right)$

  $I_{peak}=7.94\text{A}$

Formula used:

Expression for peak magnetic energy stored in inductor is

  $U_{peak}=\frac{1}{2}L{I}^2{}_{peak}$.......... (1)

Expression for average magnetic energy stored in inductor is

  $U_{avg}=\frac{1}{4}L{I}^2{}_{peak}$.......... (2)

Here, $L$ is inductance of inductor and $I_{peak}$ is the peak current.

Calculation:

Substitute $36\times {10}^{-3}\text{H}$ for $L$ and $7.94\text{A}$ for $I_{peak}$ in equation (1)

  $\begin{array}{l}{U}_{peak}=\frac{1}{2}\times \left(36\times {10}^{-3}\right)\times {\left(7.94\right)}^2\\ \boxed{U_{peak}=1.1\text{J}}\end{array}$

Average energy is calculated as

  $\begin{array}{l}{U}_{avg}=\frac{1}{4}\times \left(36\times {10}^{-3}\right)\times {\left(7.94\right)}^2\\ \boxed{U_{avg}=0.57\text{J}}\end{array}$

Conclusion:

Hence, the required peak and average magnetic energy is $\boxed{U_{peak}=1.1\text{J}}$ and $\boxed{U_{avg}=0.57\text{J}}$