Chapter 35, Problem 56A

(a)

To determine

The current drawn by the night light.

Answer to Problem 56A

The current drawn by the night light bulb is $0.033\text{A}$ .

Explanation of Solution

Given info:

Power of the night light bulb, $P=4\text{W}$

Supply voltage for the night light bulb, $V=120\text{V}$

Formula Used:

Write the expression to calculate the current drawn by the night light bulb.

  $I=\frac{P}{V}$

Calculation:

Substitute given values in the above expression.

  $\begin{array}{c}I=\frac{4\text{W}}{120\text{V}}\\ =0.033\text{A}\end{array}$

Conclusion:

The current drawn by the night light bulb is $0.033\text{A}$ .

(b)

To determine

The resistance of the nightlight filament.

Answer to Problem 56A

The resistance of the night light bulb is $3636.4\Omega$ .

Explanation of Solution

Given info:

Current drawn by the night light bulb is $0.033\text{A}$ .

Supply voltage for the night light bulb is, $V=120\text{V}$

Formula Used:

Write the expression to calculate the resistance of the filament.

  $R=\frac{V}{I}$

Calculation:

Substitute given values in the above expression.

  $\begin{array}{c}R=\frac{120\text{V}}{0.033\text{A}}\\ =3636.4\Omega \end{array}$

Conclusion:

The resistance of the night light bulb is $3636.4\Omega$ .

(c)

To determine

The energy uses by the night light in one month.

Answer to Problem 56A

The energy used by the night light in one month is $2.88\text{kWh}$ .

Explanation of Solution

Given info:

Power of the night light bulb, $P=4\text{W}$

Number of days, $n=30\text{day}$

Formula Used:

Write the expression to calculate the energy used by the night light.

  $E=\frac{Pnt}{1000}$

Calculation:

Substitute values in the above expression.

  $\begin{array}{c}E=\frac{\left(30\text{day}\right)\left(\frac{24\text{h}}{1\text{day}}\right)\left(4\text{W}\right)}{1000}\\ =2.88\text{kWh}\end{array}$

Conclusion:

The energy used by the night light in one month is

  $2.88\text{kWh}$ .

(d)

To determine

The cost of operation for the month.

Answer to Problem 56A

The cost of operation of night light in a month is $28.8\text{cent}$ .

Explanation of Solution

Given info:

Utility rate of the electricity, $r=10\text{cent}/\text{kWh}$

Formula Used:

Write the expression to calculate the cost of operation.

  $C=E\cdot r$

Calculation:

Substitute values in the above expression.

  $\begin{array}{c}C=\left(2.88\text{kWh}\right)\left(10\text{cent}/\text{kWh}\right)\\ =28.8\text{cent}\end{array}$

Conclusion:

The cost of operation of night light in a month is $28.8\text{cent}$ .