Example 7.4: Energy in DC circuits

Question: A 150W light bulb is connected to a 120V line. What is the current drawn from the line? What is the resistance of the light bulb whilst it is burning? How much energy is consumed if the light is kept on for 6 hours? What is the cost of this energy at 8 cents/kWh?
 
Answer: Since power is equal to $I\,V$, it follows that

$$I = \frac{P}{V} = \frac{(150)}{(120)} = 1.25\,{\rm A}.$$

From Ohm's law, the resistance of the light bulb is

$$R = \frac{V}{I} = \frac{(120)}{(1.25)} = 96 \,\Omega.$$

The energy $W$ consumed is the product of the power $P$ (the energy consumed per unit time) and the time period $t$ for which the light is on, so

$$W = P\,t = (150)\,(6)\,(60)\,(60) = 3.24\times 10^6\,{\rm J}.$$

Since, $1\,{\rm kWh} \equiv 3.6\times 10^6\,{\rm J}$, it follows that

$$W = \frac{(3.24\times 10^6)}{(3.6\times 10^6)} = 0.9\,{\rm kWh}.$$

The cost $c$ of the electricity is product of the number of kilowatthours used and the cost per kilowatthour, so

$$c = (0.9)\,(0.08) = 0.072\,\,{\rm dollars} = 7.2\,\,{\rm cents}.$$