Energy and Power

As the electric charges pass through a circuit, they carry energy and distribute this electrical energy to the different loads they pass through. When these electric charges is pushed by an electric potential, V, and produces a current,I for a time, t, the energy they distribute may be expressed as

$$E=VIt\quad\quad\quad\quad\quad (1)$$.

The amount of electrical energy used  or supplied per unit of time is called Power and this is given by the equation

$$P=\frac Et\quad\quad\quad\quad\quad (2)$$.

The power may be expressed in terms of the potential difference and current. To get this, we have to substitute (1) to (2), that is

$$P=\frac Et=\frac {VIt}{t}=VI\quad\quad (3)$$.

As the charge pass through the circuit and encounter a load with resistance, R, the power can be computed using the Ohm's Law.

Using the relationship $$V=IR$$, the equation will become

$$P=VI\\ \;\;\,=(IR)(I)\\ \;\;\,P=I^2R\quad\quad\quad\quad(4)$$

Substituting the equation $$I=\frac VR$$ to (3), the formula for P is

$$P=\frac VR\\ \;\;\,=(V)(\frac VR)\\ \;\;\,P=V^2R\quad\quad\quad\quad(5)$$.

The unit for Power is Watt (W).

Example 1.

A 100-W bulb has an electric potential of 120 V. What is its current?

Given:

$$P=100\;W\\ V=120\;V$$

Solution:

To solve for the current, we use formula (3) and derive the formula. Hence, current can be calculated as

$$I=\frac PV=\frac {100\;W}{120\;V}=0.83\;A$$.

Example 2.

Calculate for the resistance of a 1,491-W airconditioner which uses a current of 20 A.

Given:

$$P=1,491\;W\\ I=20\;A$$

Solution:

To solve this problem, formula (4) is used. The resistance of the airconditioner is given by

$$R=\frac{P}{I^2}=\frac{1,491\;W}{(20\;A)^2}==\frac{1491\;W}{400\;A^2}=3.73 \;\Omega$$.

Example 3.

A washing machine with a resistance of $$25\;\Omega$$ can work as it is connected to a 220-V power supply. Solve for the machine's power.

Given:

$$R=25\;\Omega\\ V=220\;V$$.

Using formula (5), the rate at which the machine uses energy is

$$P=\frac{V^2}{R}=\frac {(220\;V)^2}{25\;\Omega}=\frac{48,400\;V^2}{25\;\Omega}=1,936\;W$$.