**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,

**for a time,**

*I***, the energy they distribute may be expressed as**

*t*\(E=VIt\quad\quad\quad\quad\quad (1)\).

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

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