 # Electrical energy and power

## Energy

When electrons start to travel around a circuit, there is a conversion of energy from the chemical energy of the battery to electrical energy. This energy is used to 'push' the electrons through loads.

For example, the electrons may excite the filament of a lamp or a kettle to glow and give out heat and light. Or they might drive a motor, or sound a buzzer.

Electrical energy E is the power P times the time t the power is applied for:

\$\$E = Pt\$\$

The unit of electrical energy is joule (J).

## Power

Electrical power P is the voltage V times the current I:

\$\$P = IV\$\$

The unit of electrical power is watt (W), and is equivalent to \$1 J/s\$, one joule of energy per second.

A kilowatt (kW) is the power level which consumes 1 kJ (one thousand joules). A kW-s is the power that consumes 1 kJ in one second. A kWh is therefore the power that consumes 3600 joules in one hour. (There are 3600 seconds in one hour)

### Example Ammeter and voltmeter in a circuit

In the photograph, it can be seen that LEDs are being powered by a power supply. The current is 49 mA, and the voltage is 5V.

The power P used to light the LEDs is:

\$\$P = IV = 49 x 10^{-3} ⋅ 5 \$\$ \$\$= 0.245 W\$\$

Question: if the LEDs are left on for one hour (3600 s), how much energy would be consumed?

The energy consumed is:

\$\$E = Pt = 0.245 ⋅ 3600 \$\$ \$\$= 882 J\$\$

### Power Dissipation

When power is used to push electrons through a resistor, we say the power is 'dissipated'. This is another way of saying that the energy the power is supplying is being converted to heat, light or mechanical energy, and lost from the electrical energy of the circuit.

Since Ohm's Law tells us that \$V = IR\$, we can add this to the power equation \$P = IV\$ to get:

\$\$P = IV = I⋅IR = I^2R\$\$

### Example

How much more power is dissipated in a resistor if the current is doubled?

Let \$I_1\$ be the first case, and \$I_2\$ the second. R is the same in both cases.

\$\$P_1 = I_1^2R\$\$ \$\$P_2 = I_2^2R\$\$

Since \$I_2 = 2I_1\$,

\$\$P_2 = (2I_1)^2R = 4I_1^2R = 4P_1\$\$

To double the current, four times as much power must be dissipated!

The reason for this is that the resistance of a load makes it hard to increase the current, so it is not linear.

Demonstration of the electric motor principle. A cork with a copper wire wrapped tightly around it, on a brass rod, suspended between two flat magnets.

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