Chapter 21, Problem 38QAP

To determine

(a)

The rms current passing through the resistor.

Answer to Problem 38QAP

The rms current is $\text{0}\text{.353}\text{A}$.

Explanation of Solution

Given info:

The $V_{\max }$ voltage is $\text{50}\text{.0 V}$.

The resistance of the resistor is $100.0\Omega$.

Formula used:

The relation between rms current and r.m.s. voltage is,

  $I_{rms}=\frac{V_{rms}}{R}$  Equation-1

Here, ${\text{V}}_{\text{rms}}$ is the V r.m.s. voltage, $I_{\text{rms}}$ is the rms current and $R$ is the resistor.

The relation between V r.m.s. voltage and peak voltage is,

  $V_{rms}=\frac{V_{\max }}{\sqrt{2}}$  Equation-2

Here, ${\text{V}}_{\text{rms}}$ is the V r.m.s. voltage and $V_{\max }$ is the maximum voltage drop is,

Using equation-1 the expression for peak voltage is

  $I_{rms}=\frac{V_{\max }}{\sqrt{2}R}$   Equation-3

Calculation:

Finding rms current:

Substitute the all given value in equation-3 to find $V_{\max }$,

  $\begin{array}{l}{I}_{rms}=\frac{50.0\text{V}}{\sqrt{2}\times 100}\\ =0.353\text{A}(rms)\text{Equation-4}\end{array}$

Conclusion:

From equation-4 the rms current is $\text{0}\text{.353}\text{A}$.

To determine

(b)

The maximum voltage across the resistor.

Answer to Problem 38QAP

The maximum voltage is $\text{707}\text{V}$.

Explanation of Solution

Given info:

The rms current is $\text{2}\text{.50}\text{A}$.

The resistance of the resistor is $200.0\Omega$.

Formula used:

The relation between rms current and maximum voltage is,

  $I_{rms}=\frac{V_{\max }}{\sqrt{2}R}$

Here, ${\text{V}}_{\max }$ is the maximum voltage, $I_{\text{rms}}$ is the rms current and $R$ is the resistor.

Rearrange the equation to find the $V_{\max }$

  $V_{\max }=\sqrt{2}{I}_{rms}R$   Equation-4

Calculation:

Finding maximum voltage:

Substitute the all given value in equation-4 to find $V_{\max }$,

  $\begin{array}{c}{V}_{\max }=\sqrt{2}\times 2.5\text{A}\times 200\Omega \\ =707\text{V}\text{Equation-5}\end{array}$

Conclusion:

From equation-5 the maximum voltage is $\text{707}\text{V}$.

To determine

(c)

The resistance of the resistor.

Answer to Problem 38QAP

The resistance of the resistor is $\text{156}\Omega$.

Explanation of Solution

Given info:

• The rms current is $\text{127}m\text{A}$.

Converting milli-Ampere to ampere.

  $\begin{array}{c}\text{127}\text{mA}\text{=}\text{127}\text{mA}\times \frac{1\text{A}}{1000\text{mA}}\\ =127\times {10}^{-3}\text{A}\end{array}$

• The maximum voltage is $\text{28}\text{.0}\text{V}$.

Formula used:

The relation between rms current and maximum voltage is,

  $I_{rms}=\frac{V_{\max }}{\sqrt{2}R}$

Here, ${\text{V}}_{\max }$ is the maximum voltage, $I_{\text{rms}}$ is the rms current and $R$ is the resistor.

Rearrange the equation to find the $R$.

  $R=\frac{V_{\max }}{\sqrt{2}{I}_{rms}}$   Equation-6

Calculation:

Finding maximum voltage:

Substitute the all given value in equation-6 to find $R$,

  $\begin{array}{c}R=\frac{28.0\text{V}}{\sqrt{2}\times 1127\times {10}^{-3}\text{A}}\\ =156\Omega \text{Equation-7}\end{array}$

Conclusion:

From equation-7 the resistance of the resistor is $\text{156}\Omega$.