Power

4.6. Power#

The electric potential difference, or voltage $V$ , between two points is work done in moving a charge from one point to the other.

V = \frac{d w}{d q}

By definition the rate at which work is done is power, $P$ .

P = \frac{d w}{d t}

Using the definition of instantaneous voltage, and rearranging to solve for the change in work $d w$ :

d w = v d q

The change in work, $d w$ per unit time $d t$ is therefore:

\frac{d w}{d t} = v \frac{d q}{d t}

But remember, the definition of current is $i = \frac{d q}{d t}$ , thus Power can be written as:

P = v i

For simpler DC-steady state circuits:

P = V I

Units of Power#

Power is equal to voltage times current.

P = V I

SI units for voltage $V$ are $[\frac{J}{C}]$ and the SI units for current $I$ are $[\frac{C}{s}]$ .

P = [\frac{J}{C}] [\frac{C}{s}]

Canceling the units coulombs, $C$ , power $P$ ends up in units of $[\frac{J}{S}]$ . One joule ( $J$ ) per second ( $s$ ) is equal to one watt ( $W$ ).

P = [\frac{J}{s}] = [W]

Power, Current, Voltage and Resistance#

When current flows through a resistor, the absorbed electrical energy is dissipated as thermal energy.

The rate at which this occurs is referred to as power dissipation.

P = V I

Using Ohm’s law, two alternate forms relating power to current, resistance, and voltage can be created using

Ohm’s law as $V = I R$

P = I^{2} R

Ohm’s law as $I = \frac{V}{R}$

P = \frac{V^{2}}{R}

The image below shows how power, voltage, current and resistance can all be related.

power voltage current resistance wheel Image Credit: WireCharm, Creative Commons Attribution 3.0 License

Worked Example

GIVEN:

The circuit diagram below:

circuit with battery and lamp

Image Credit: Applied Industrial Electricity, Design Science License

In the above circuit, we know we have a battery voltage of 18 volts and a lamp resistance of 3 $Ω$ .

FIND:

The electrical power disspated by the lap, $P_{l}$

SOLUTION:

In the above circuit, we know we have a battery voltage of 18 volts and a lamp resistance of 3 $Ω$ . Using Ohm’s Law to determine current, we get:

I = \frac{V}{R} = \frac{18 V}{3 Ω} = 6 A

Now that we know the current, we can take that value and multiply it by the voltage to determine power:

P = V I = (18 V) (6 A) = 108 W

This tells us that the lamp is dissipating (releasing) 108 watts of power, most likely in the form of both light and heat.

P_{l} = 108 W

Power

Contents

4.6. Power#

Units of Power#

Power, Current, Voltage and Resistance#