Generalized Power Law

In electrical engineering, the Power Law (also called the Generalized Power Law or Watt’s Law) is the fundamental relationship between energy transfer, potential difference, and charge flow.

The Generalized Power Law Equation

The Power Law states that electrical power (P) is the product of voltage (V) and current (I). We multiply voltage by current to calculate electric power (P) measured in Watts (W). This gives us the equation:

/[ P = VI /]

• P (Electric Power) — Measured in Watts (W), this is the rate at which energy is converted into another form, or the work that is being performed.
• V (Voltage) — Measured in Volts (V), this is the “pressure” or potential difference.
• I (Current) — Measured in Amps (A), this is the rate of flow of electric charge.

This equation is “generalized” because it is a law of conservation of energy. It works for any electrical device (motors, batteries, LEDs, heaters) whether it follows Ohm’s Law or not. [1]

The Power Triangle

Much like the Ohm’s Law triangle (V = IR) the Generalized Power Law is often visualized as a triangle to make the relationship between each variable even more clear, and assist with fast calculations.

All you do is isolate the value you are calculating for, and then the position of the other two tells the relationship, and which operation (multiplication or division) that you use.

• Calculate Power (P) — V multiplied by I.
• Calculate Current (I) — P divided by V.
• Calculate Voltage (V) — P divided by I.

These are all the various forms that the Power Law can take.

\[ P = VI /]

\[ V = \frac{P}{I} \]

\[ I = \frac{P}{V} \]

We can add even more forms to the Power Law by substituting in Ohm’s Law for unknown variables, which we see below.

Combining with Ohm’s Law — Ohm’s Law Substitution into the Power Law

When you are dealing specifically with resistors (components that follow Ohm’s Law, V = IR) you can substitute V or I to get two other famous versions of the power law. These are often used to calculate “heat loss” in wires.

Substitution for V. When we don’t know the voltage, but know the resistance and can measure the current.

P = (IR) \times I \rightarrow P = I^2 R

Best for: Finding power when you know the current (common in series circuits).

Substitution for I.

P = V \times (\frac{V}{R}) \rightarrow P = \frac{V^2}{R}

Best for: Finding power when you know the voltage (common in parallel household circuits).

When To Use Each Law

Every variation of the Generalized Power Law has it’s own practical uses, performing different functions, which is summarized in the list below.

• Generalized Power Law (P = VI) — Always works for all electrical components. |
• Joule’s Law (Heating) (P = I^2 R) — Calculating heat loss in wires/cables.
• Voltage-Resistance Power (P = V²/R) — Calculating power for devices plugged into a constant voltage outlet.

Power Law Examples

(As usual, try to do the question yourself first, before reading the explanation and looking up the answer!)

Basic Questions

Imagine a 240-V generator that supplies 50A of electric current to a circuit. This profile can fit that required for an oven, or powerful EV charger. How much power is being drawn into the circuit?

240V is one of the standard voltages used in North America (the other being 120V). 50A is also a realistic current for an oven, powerful EV charger. We simply use the Power Law to calculate how much power these devices require.

\[ P = VI = (240V)(50A) = 12,000 W = 12 kW \]

In North America, an oven might draw 12 kW of power at full-capacity. (In order to calculate home much energy is required to power the oven — and how much that costs — we will use the concept of kilowatt-hours later on.)

Example III — Ohm’s Law Substitution

We have a circuit with a current running through it, running through a $200-\Omega$ resistor where we require 0.30 A of current. Find the power rating of the resistor.

Since we already know I and R, we can use the Power Law (P=VI) and substitute in Ohm’s Law (V=IR) for the missing voltage (V) term. We arrive at the expression

\[ P = I^2 R = (0.30A)^2 (200\Omega) = 1.8 W \]

To prevent a resistor from burning out, we want the power rating of any resistor to be twice the wattage calculated in the power equation. [2] So in this case, we might want to use a 4W power rating resistor, to be safe.

We have a circuit with a current running through it, running through a $100-\Omega$ resistor where we require 0.20 A of current. Find the power rating of the resistor. [2]

\[ P = I^2 R = (0.20A)^2 \times 100-\Omega = 4 W \]

To prevent the resistor from burning out, to be safe we want to double that and use one with a power rating of at least 8 W.

How many kilowatts of power are delivered to a circuit by a 240-V generator, supplying 20A to the circuit? [2]

\[ P = VI = (240-V)(20-A) = 4800 W = 4.8 kW \]

4.8 kW of power are delivered to a circuit by a 240-V generator with a current of 20A.

If a toaster draws 1500 W of power when toasting 2 slices of bread, in a standard North American home running on 120-V for small appliances, what current can we expect going into the device when its operating?

For this problem, we use the Power Law (P=VI) to solve for the current.

\[ I = \frac{P}{V} = \frac{1500 W}{120 V} = 12.5 A \]

A toaster running at max power might draw 12.5 A of current.

Now, imagine a toaster that draws 15A of current from the grid with a voltage of 120-V. Using Joules Law, what is the Resistance (R) required to get maybe 1200 W of heat from the wires in the toaster?

\[ P = I^2 R \]

\[ R = \frac{ P }{I^2} = \frac{1200W}{(15A)^2} = 5.33 \Omega \]

We want a resistance in our resistors of about 5.33 Ohms ($5.3 \Omega$), but to be safe (i.e to ensure that they can withstand the heat of cooking) we might want a total resistance of at least $12 \Omega$.

Heat Dish Example

If we have a heat-dish rated at 1500 W, that is plugged in a wall socket in North America (with a standard voltage of 120V) what is the current drawn by that heat-dish at full power?

/begin{equation} I & = \frac{P}{V} // & = \frac{1500W}{120V} // & = 12.5 A /end{equation}

We can see that the heat-dish draws approximately 12.5 A of current through its coils, which heats them up, producing a red-orange light and thermal radiation (heat) that it bounces off the dish.

Image that we have a space-heater rated to 1500W, however it has two switches, one at 600W and another at 900W, with a dial that goes from 0-6. We have the 900W switch engaged and the dial full-blast (set to 6), and running on 120V of power. What is the amperage of the current drawn?

In this question, out we are drawing 900W of power with a voltage of 120V. Therefore, we are using the Generalized Power Law for Current (Amps, A). Therefore we rearrange the equation:

\[ P = VI \rightarrow I = \frac{P}{V} \]

We then solve the equation

\[ I = \frac{900W}{120V} = 7.5 A \]

So our coil space heater draws about 7.5 A of electric current from the grid in order to run, each second.

Further Reading

• Generalized Power Law Problem Set I [beta]