Q49P - Fundamentals Of Physics

Question

A 100 W lightbulb is plugged into a standard 120V outlet. (a) How much does it cost per 31-day month to leave the light turned on continuously? Assume electrical energy costs US $ 0.06/kWh. (b) What is the resistance of the bulb? (c) What is the current in the bulb?

Step-by-Step Solution

Answer

1. The cost per 31-day month to leave the light turned on continuously is US $4.46.
2. The resistance of the bulb is $144\mathrm{\Omega }$. 
3. The current in the bulb is 0.833A

Step 1: Identification of given data

1. The power of the bulb is P = 100W
2. The potential difference is V = 120V
3. Time the light is turned on, t = 31-day month
4. The electrical energy cost is US $ 0.06.kW.h 

Step 2: Significance of rate of energy dissipation

The power or rate of energy dissipation, in an electrical device across which a potential difference is equal to the product of current and potential difference. It can also be defined as energy transferred per unit time.

 

Here, we need to find the amount of electric energy utilized to turn on a light bulb continuously for a month. Then, we can multiply that value to get the monthly cost. The resistance and current through the bulb can be calculated using the equations of power.

 

Formulae:

The energy transferred in the system, E = Pt                                                            …(i)

Here, P is the power, E is electric energy, t is the time.

From Eq. 26-28, the resistance due to power is given by, $R=\frac{V^2}{P}$                             …(ii)

Here, P is the power, R is resistance, V is the potential difference.

 

From Eq. 26-26, the current flowing is given by,  $i=\frac{P}{V}$                                            …(iii)

Here, P is the power, i is current, V is the potential difference.

Step 3: (a) Determining the cost to leave the light turned on

Let us assume a 31-day month, so the electric energy required to keep 100W bulb turned on continuously is given using equation (i) as follows:

$$\begin{gathered} \mathrm{E}=100\mathrm{W}\times \frac{31 \mathrm{day}}{\mathrm{month}}\times \frac{1 \mathrm{kW}}{{10}^3\mathrm{W}}\times \frac{24\mathrm{hour}}{1\mathrm{day}} \\ =74.\mathrm{r}4\mathrm{kW}.\mathrm{h} \end{gathered}$$

Now, the cost of electric energy is given as: US $0.06/kW . h

So, the monthly cost will become can be given as follows:

$$\begin{gathered} \mathrm{monthly} \mathrm{cost}=74.4\mathrm{kW}.\mathrm{h}\times \frac{\mathrm{US} \$0.06}{\mathrm{kW}.\mathrm{h}} \\ =\mathrm{US} \$4.46 \end{gathered}$$

Hence, the value of the cost is US $ 4.46.

Step 4: (b) Determining the resistance

Using the given data in equation (ii), we can get the value of the resistance of the bulb as follows:

$$\begin{gathered} R=\frac{{\left(120V\right)}^2}{100W} \\ =144\Omega \end{gathered}$$

Hence, the value of the resistance is $144\Omega$.

Step 5: (c) Determining the current value

Using the given data in equation (iii), we can get the current value in the bulb as follows:

$$\begin{gathered} i=\frac{100W}{120V} \\ =0.833A \end{gathered}$$

Hence, the value of the current is 0.833A.