- Burn a $40\, W$ light bulb for $24\, h$ straight.
- Operate an electric oven for $6.0\, h$ if it carries a current of $20.0\, A$ at $220\, V$.
This question aims to find the cost of electrical energy for a $40\, W$ light bulb and an electric oven carrying the current of $20.0\, A$ at $220\, V$ in the given time.
The movement of electric charges generates kinetic energy usually known as electric energy. The flow rate of charges, which is the fast movement, ensures the carriage of more electrical energy. The electrical circuit delivers the electric potential and electric current both of which combines and results in the supply of electrical energy. More generally, this energy can be regarded as the transformed form of potential and electrical energy. The motion of electrons from one point to the other also generates electrical energy. The motion of charged particles through a media is said to be electricity or a current.
Examples of electrical energy from daily life are a thunderstorm, in which the energy is transmitted as lightning, in a power plant the storage of energy in the electrical generator which is later on transferred through wires. In addition to this, a capacitor holds the energy which can be transferred to move a current via an electrical circuit.
Expert Answer
Given the cost of electricity is:
$\$0.12\,/kW\,h$
For the light bulb:
Let $C$ be the total cost, $P$ be the power, $h$ be the time used, and $C_p$ be the cost per kilowatt-hour, then the total cost can be formulated as:
$C=P\times h \times C_p$
Now, $P=40\,W=0.040\,kW$
Therefore, $C=0.040\,kW\times 24\, h\times \$0.12\,/kW\,h$
$C=\$ 0.12$ or $C=11.52$ cents.
To operate an electric oven:
Let $P$ be the power, $I$ be the electric current and $V$ be the voltage in the circuit then:
$P=VI$
Here, $I=20.0 \,A$ and $V=220\,V$ so that:
$P=220\times 20.0$
$P=4400\,W=4.4\,kW$
Again using the cost formula:
$C=P\times h \times C_p$
$C=4.4\,kW\times 6.0\,h\times \$0.12\,/kW\,h$
$C=\$ 3.7$ or $316.8$ cents.
Example
Find the amount of energy required by a $335\,V-6\,A$ electric bulb if it is used for $25$ minutes.
Solution
Given that:
Voltage $=V=335\, V$
Current $=I=6\,A$
Time $=t=25$ minutes
$t=25\times 60=1500$ s
Now, find the power as follows:
$P=VI$
$P=(335\, V)(6\,A)=2010\,W$
Or $P=2010\,J/s$
Electrical Energy $=P\times t$
$=2010\,J/s\times 1500\,s$
Electrical Energy $=3,015,000\,J$ or $3,015\,kJ$