Chapter 22.2, Problem 40SSC

a.

To determine

Resistance of the toaster.

Answer to Problem 40SSC

  $R=28.57\Omega$

Explanation of Solution

Given:

The given circuit is shown below.

  

Source voltage, $V=120\text{V}$

Circuit current, $I=4.2\text{A}$

Formula Used:

Ohm’s law is given as,

  $V=IR$

Where, $V$ is applied voltage, $I$ is current and $R$ is resistance.

Calculation:

On substituting the values in above relation,

  $\begin{array}{c}120=4.2\times R\\ R=\frac{120}{4.2}\Omega \\ =28.57\Omega \end{array}$

Conclusion:

Therefore, the resistance of the toaster is $R=28.57\Omega$

b.

To determine

Rate of energy transformation.

Answer to Problem 40SSC

  $P=496.55\text{J/s or }496.55\text{W}$ .

Explanation of Solution

Given:

The given circuit is shown below.

  

Source voltage, $V=120\text{V}$

Circuit current, $I=4.2\text{A}$

Formula Used:

Power dissipation in resistor is given by,

  $P=\frac{V^2}{R}$

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

Calculation:

The value of resistor is $R=28.57\Omega$ from part (a).

So, the rate of transforming energy that is power is

  $\begin{array}{c}P=\frac{V^2}{R}\\ =\frac{{120}^2}{28.57}\\ =496.55\text{J/s or }496.55\text{W}\end{array}$

Conclusion:

Therefore, the rate of transforming energy is $P=496.55\text{J/s or }496.55\text{W}$