Chapter 23, Problem 92A

To determine

Maximum safe voltage.

Answer to Problem 92A

  $V=69.3\text{V}$

Explanation of Solution

Given:

The given figure is,

  

Rated power of each resistor is 5 W.

Formula Used:

Power in resistor is,

  $P=I^{2}R$

Ohm’s law is,

  $V=IR$

Calculation:

As the given circuit is a series circuit so, the current is same in all resistors.

And the largest resistor will develop most power.

  $\begin{array}{l}P=I^{2}R\\ \Rightarrow I=\sqrt{\frac{P}{R}}\\ =\sqrt{\frac{5}{220}}=0.151\text{A}\end{array}$

The total resistance is,

  $\begin{array}{l}{R}_{\text{Total}}=92\Omega +150\Omega +220\Omega \\ =462\Omega \end{array}$

So, the maximum safe voltage is

  $\begin{array}{c}V=I{R}_{\text{Total}}\\ =0.151\times 462\\ =69.3\text{V}\end{array}$

Conclusion:

Therefore, the maximum safe voltage is $V=69.3\text{V}$ .