Chapter 22, Problem 82A

a.

To determine

The cost per kWh to operate the radio.

Answer to Problem 82A

Explanation of Solution

Given:

The potential supplied by the battery, $V=9.0\text{V}$

The current through the circuit, $I=50.0\text{mA}$

The cost of the battery, $C_b=\$2.49$

The time for which the battery works, $t=300.0\text{h}$

Formula used:

The energy $\left(E\right)$ delivered during time $t$ is given by,

  $\begin{array}{c}E=Pt\\ =VIt\end{array}$

Where,

I is the current and V is the potential difference.

Calculation:

Using the above formula, the energy delivered by the battery,

  $\begin{array}{c}E=VIt\\ =\left(9.0\right)\left(50\times {10}^{-3}\right)\left(300\right)\\ =0.135\text{kWh}\end{array}$

Then the cost per kWh to operate the radio will be,

  $\begin{array}{c}C\text{'}=\frac{C_b}{E}\\ =\frac{\$2.49}{0.135}\\ =\$18.44\text{per}\text{kWh}\end{array}$

Conclusion:

Hence, the cost per kWh to operate the radio is $\$18.44\text{per}\text{kWh}$ .

b.

To determine

The new cost per kWh to operate the radio if cost is $\$0.12\text{per kWh}$ .

Answer to Problem 82A

  $\$0.02$

Explanation of Solution

Given:

The potential supplied by the battery, $V=9.0\text{V}$

The current through the circuit, $I=50.0\text{mA}$

The cost, $C\text{'}=\$0.12\text{per kWh}$

The time for which the battery works, $t=300.0\text{h}$

Formula used:

The energy $\left(E\right)$ delivered during time $t$ is given by,

  $\begin{array}{c}E=Pt\\ =VIt\end{array}$

Where,

  $\begin{array}{l}I=\text{current through circuit}\\ V=\text{ potential difference applied}\end{array}$

Calculation:

Using the above formula, the energy delivered by the battery,

  $\begin{array}{c}E=VIt\\ =\left(9.0\right)\left(50\times {10}^{-3}\right)\left(300\right)\\ =0.135\text{kWh}\end{array}$

Then the cost to operate the radio will be,

  $\begin{array}{c}{C}_t=C\text{'}\times E\\ =\$0.12\times 0.135\\ \approx \$0.02\end{array}$

Conclusion:

Hence, the cost to operate the radio is now $\$0.02$ .