Chapter 22.2, Problem 32PP

a.

To determine

Current drawn by clock.

Answer to Problem 32PP

  $I=9.59\times {10}^{-3}\text{A}$

Explanation of Solution

Given:

Resistance of clock, $R=12000\Omega$

Source voltage, $V=115\text{V}$

Formula used:

Ohm’s law is given as,

  $V=IR$

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

Calculation:

On substituting the given values in above relation

  $\begin{array}{l}115=I\times 12000\\ \Rightarrow I=\frac{115}{12000}=9.59\times {10}^{-3}\text{A}\end{array}$

Conclusion:

Therefore, the current drawn by clock is $I=9.59\times {10}^{-3}\text{A}$ .

b.

To determine

Power used.

Answer to Problem 32PP

  $P=1.1\text{W}$

Explanation of Solution

Given:

Resistance of clock, $R=12000\Omega$

Source voltage, $V=115\text{V}$

Formula used:

Power consumed is given as,

  $P=VI$

Where, $V$ is voltage and $I$ is current.

Calculation:

From previous subpart $I=9.59\times {10}^{-3}\text{A}$

So, on substituting the values in above relation

  $\begin{array}{c}P=115\times 9.59\times {10}^{-3}\\ =1.1\text{W}\end{array}$

Conclusion:

Therefore, the power used is $P=1.1\text{W}$ .

c.

To determine

Cost of operation for given time and rate.

Answer to Problem 32PP

Operation cost is $\$0.095$

Explanation of Solution

Given:

Resistance of clock, $R=12000\Omega$

Source voltage, $V=115\text{V}$

Time is $t=30\text{days}$

Rate of electricity is $ 0.12/kWh

Formula used:

The cost is given as,

  $\text{Cost = kWh}\times \text{rate}$

Calculation:

As from previous subpart, the power is $P=1.1\text{W}=1.1\times {10}^{-3}\text{kW}$

And time in hour is $t=30\text{days}=\left(30\times 24\right)\text{hours}=720\text{hours}$

So, the cost of electricity is

  $\text{Cost = (1}\text{.1}\times {\text{10}}^{-3}\times 720)\times \$0.12=\$0.095$

Conclusion:

Therefore, the cost of electricity is $\$0.095$ .