Chapter 22, Problem 79A

a.

To determine

The maximum safe current if the maximum safe power is $5.0\times {10}^1\text{W}$ .

Answer to Problem 79A

  $1.12\text{A}$

Explanation of Solution

Given:

The resistance in the circuit, $R=40.0\Omega$

Formula used:

The electric power $\left(P\right)$ across a circuit is given by,

  $\begin{array}{c}P=IV\\ =I^{2}R\end{array}$

Where, I is the current through circuit, V is the potential difference applied and R is the resistance.

Calculation:

Using the above formula and the given values, the maximum safe current will be,

  $\begin{array}{c}P=I^{2}R\\ 50=I^2\left(40\right)\\ I=1.12\text{A}\end{array}$

Conclusion:

Hence, thethe maximum safe current is $1.12\text{A}$ .

b.

To determine

The maximum safe voltage if the maximum safe power is $5.0\times {10}^1\text{W}$ .

Answer to Problem 79A

  $44.72\text{V}$

Explanation of Solution

Given:

The resistance in the circuit, $R=40.0\Omega$

Formula used:

The electric power $\left(P\right)$ across a circuit is given by,

  $\begin{array}{c}P=IV\\ =\frac{V^2}{R}\end{array}$

Where, I is the current through circuit, V is the potential difference applied and R is the resistance.

Calculation:

Using the above formula and the given values, the maximum safe voltage will be,

  $\begin{array}{c}P=\frac{V^2}{R}\\ 50=\frac{V^2}{40}\\ V=44.72\text{V}\end{array}$

Conclusion:

Hence, themaximum safe voltage is $44.72\text{V}$ .