Chapter 18, Problem 46P

(a)

To determine

The peak voltage.

Answer to Problem 46P

  $933volts$

Explanation of Solution

Given info:

Root mean square Voltage, $V_{rms}=660V$

Power, $P=1800W$

Formula used:

Formula to calculate peak voltage:

  $V_0=\sqrt{2}{V}_{rms}$

Where, $V_0$ is peak voltage and $V_{rms}$ is root mean square voltage.

Calculation:

Substitute all the given values in the expression, $V_0=\sqrt{2}{V}_{rms}$

  $V_0=\sqrt{2}\times 660=933V$

Conclusion: The peak voltage is $V_0=933V$ .

(b)

To determine

The peak current.

Answer to Problem 46P

  $3.86A$

Explanation of Solution

Given info:

The peak voltage, $V_0=933V$ .

Root mean square Voltage, $V_{rms}=660V$

Power, $P=1800W$

Formula used:

Formula of an average power:

  $\overline{P}=I_{rms}{V}_{rms}$

  $\overline{P}=\frac{1}{2}\frac{V_0^2}{R}=\frac{V_{rms}^2}{R}$

Where, $V_0$ is peak potential difference in volts, $V_{rms}$ is peak potential difference in volts, and $R$ is resistance.

Calculation:

By using the formula of average power,

  $\overline{P}=I_{rms}{V}_{rms}$

  $I_{rms}=\frac{\overline{P}}{V_{rms}}$

So, the peak current:

  $I_0=\sqrt{2}{I}_{rms}$

So, $I_{rms}=\frac{I_0}{\sqrt{2}}$

Therefore, $\frac{I_0}{\sqrt{2}}=\frac{\overline{P}}{V_{rms}}$

  $I_0=\frac{\sqrt{2}\overline{P}}{V_{rms}}$

  $I_0=\frac{\sqrt{2}\times 1800}{660}=3.86A$

Conclusion: The current, $I_0=3.86A$ .