Ohm's Law, Power and Energy in Circuits
Four Equations that Run Every Circuit
Electronics is a practical subject, but behind every circuit there are four mathematical relationships that appear on virtually every exam paper. Master these equations — and the skill of rearranging them — and calculation questions become straightforward. This lesson works through all four with worked examples you can follow step by step.
Ohm’s Law: V = IR
Ohm’s law states that voltage equals current multiplied by resistance:
$$V = IR$$
where V is voltage in volts (V), I is current in amperes (A) and R is resistance in ohms (Ω). Rearranging to make I or R the subject:
$$I = \frac{V}{R} \qquad R = \frac{V}{I}$$
To rearrange, remember the triangle: cover the quantity you want with your thumb and the remaining two show whether to multiply or divide. Always check that your answer is physically sensible — a current of 50 A through a 470 Ω resistor would be unrealistic for a battery circuit and signals a calculation error.
Ohm’s law: \(V = IR\). Rearranged: \(I = V/R\) and \(R = V/I\). V in volts, I in amperes, R in ohms.
Example 1 — find V: A 470 Ω resistor carries 20 mA. What is the voltage across it?
Convert mA to A first: 20 mA = 0.020 A. Then \(V = IR = 0.020 \times 470 = 9.4 \text{ V}\).
Example 2 — find I: A 9 V supply is connected to a 2.2 kΩ resistor. What current flows?
Convert kΩ to Ω: 2.2 kΩ = 2200 Ω. Then \(I = V/R = 9/2200 = 0.00409 \text{ A} = 4.09 \text{ mA}\).
Example 3 — find R: A voltage of 6 V drives a current of 12 mA. What is the resistance?
\(R = V/I = 6/0.012 = 500 \text{ }\Omega\)
Electrical Power: P = IV
Electrical power is the rate of energy transfer. The basic form is:
$$P = IV$$
Combining this with Ohm’s law gives two more forms that are useful when you know only current and resistance, or only voltage and resistance:
$$P = I^2 R \qquad P = \frac{V^2}{R}$$
Power is in watts (W). A component’s power rating (e.g. 0.25 W or 0.5 W for a resistor) must not be exceeded, or it will overheat. You will use \(P = I^2R\) to check whether a chosen resistor’s power rating is sufficient.
Three forms of the power equation: \(P = IV\), \(P = I^2R\), \(P = V^2/R\). Power is in watts (W).
Example 4 — P = IV: A LED circuit draws 20 mA from a 9 V supply. What is the power consumed?
\(P = IV = 0.020 \times 9 = 0.18 \text{ W}\)
Example 5 — check resistor rating: A 470 Ω resistor carries 20 mA. Is a 0.25 W resistor safe?
\(P = I^2R = (0.020)^2 \times 470 = 0.000400 \times 470 = 0.188 \text{ W}\). This is less than 0.25 W, so the resistor is safe.
Example 6 — P = V2/R: A 9 V supply is connected directly across a 100 Ω resistor. What power does it dissipate?
\(P = V^2/R = 81/100 = 0.81 \text{ W}\). A 0.25 W resistor would be destroyed; at least a 1 W rating is required.
Energy Transfer: E = Pt
Energy transfer is calculated when you know how long a device operates:
$$E = Pt$$
where E is energy in joules (J), P is power in watts (W) and t is time in seconds (s). This formula applies to time-on-load contexts: how much energy does a 0.5 W relay consume if it stays energised for 30 seconds?
\(E = Pt = 0.5 \times 30 = 15 \text{ J}\)
Take care with unit prefixes. Milliseconds must be converted to seconds (divide by 1000) and milliamperes to amperes (divide by 1000) before substituting into any equation. A calculation error caused by leaving current in mA instead of A is a common and avoidable mistake.
Linking the Equations Together
The four equations \(V = IR\), \(P = IV\), \(P = I^2R\) and \(P = V^2/R\) are not separate ideas — they are the same relationship rearranged for whichever pair of quantities is given in the question. Choosing the right form first saves both time and arithmetic. If the question gives you voltage and resistance, jump straight to \(P = V^2/R\) rather than first finding the current. If it gives you current and resistance, use \(P = I^2R\). The fastest students never compute an intermediate quantity unless they have to.
A useful sanity check is to keep one eye on the order of magnitude. A household appliance draws watts to kilowatts; a portable LED indicator draws milliwatts to tens of milliwatts; a microcontroller drawing 10 mA from 5 V dissipates \(P = 5 \times 0.01 = 0.05\) W — a tiny fraction of a watt. If your calculator shows several watts for a single resistor in a small signal circuit, you have almost certainly forgotten to convert milliamperes to amperes, or millivolts to volts. Spotting the wrong order of magnitude before writing the final answer recovers marks that would otherwise be lost to a unit slip.