Chapter 18, Problem 28P

To determine

The maximum Voltage.

Answer to Problem 28P

  $25.53V$

Explanation of Solution

Given info:

Power, $P=\frac{1}{4}W=0.25W$

Resistance, $R=2.7k\Omega =2.7\times {10}^3\Omega$

Formula used:

Expression for the resistance in terms of power and voltage:

  $R=\frac{V^2}{P}$

Where, $R$ is resistance, $P$ is power and $V$ is voltage.

Calculation:

Rearrange the expression, $R=\frac{V^2}{P}$

Therefore, a new expression, $V=\sqrt{PR}$

Substitute the given values:

  $V=\sqrt{\left(0.25W\right)\left(2.7\times {10}^3\Omega \right)}$

  $V=\sqrt{\left(0.25W\right)\left(2700\Omega \right)}=25.53V$

Conclusion: The maximum voltage is $25.53V$ .