T1 · Published 4 September 2026 · By The Ham Desk

Ohm's Law and Power: The Four Formulas Everything Else Rests On

Voltage, current, resistance and power are tied together by four short formulas. Every question in this unit can be answered with them, a pencil and thirty seconds - and every question in the exam that involves a number will need at least one of them.

Band none
Study time 16
Level Core
questions 10

Waveform and antenna diagram for Electrical Principles and Components — Electrical Principles and Components — waveform overlay and radiation pattern

Pre-flight briefing

Two laws, four rearrangements

Ohm's law states that voltage equals current times resistance: E = I x R. Reading it the other two ways gives I = E / R and R = E / I. The power law states that power equals voltage times current: P = I x E. Because E = I x R, power can also be written P = I squared x R or P = E squared / R, which is the form you want when the resistance is known and the current is not.

Units keep you honest: volts for E, amperes for I, ohms for R, watts for P. A 12 V supply delivering 2 A is delivering 24 W, and that same 2 A through a 6 ohm resistor is 12 V across it and 24 W dissipated as heat.

Series, parallel and the practical consequences

Resistances in series add: 100 + 200 + 300 ohms is 600 ohms. Two equal resistors in parallel give half the value, and the general rule is the reciprocal of the sum of reciprocals. In a station the same arithmetic explains voltage drop along a long 12 V power lead - the wire's own resistance subtracts from the voltage your radio sees - and why a dummy load rated 100 W gets hot: it is converting 100 W of RF into heat.

Terms you must be able to define
Term	What the examiner wants to hear
Ohm's law	E = I x R; the relationship between voltage, current and resistance.
Power law	P = I x E; combined with Ohm's law it gives P = I^2 R and P = E^2 / R.
Parallel resistors	Two equal values in parallel give half; unequal values follow the reciprocal rule.
Voltage drop	The loss of voltage along a conductor caused by its own resistance carrying current.

Question deck 10 questions

T2A01 Recall

What is the voltage across a 100 ohm resistor carrying 0.5 amperes?

T2A02 Recall

How much power is consumed by a transmitter drawing 2 amperes from a 12 volt supply?

T2A03 Core

A circuit has 12 volts across it and draws 0.5 amperes. What is its resistance?

T2A04 Core

An amplifier delivers 100 watts into a 50 ohm load. What is the current in the load?

T2A05 Recall

What is the unit of electrical resistance?

T2A06 Core

Three resistors of 100, 200 and 300 ohms are connected in series. What is the total resistance?

T2A07 Core

Two 100 ohm resistors are connected in parallel. What is the combined resistance?

T2A08 Stretch

A 24 ohm resistor has 12 volts across it. How much power does it dissipate?

T2A09 Stretch

A 12 volt battery is connected to a 3 ohm load through leads whose total resistance is 1 ohm. What current flows?

T2A10 Stretch

Which formula gives power when voltage and resistance are known but current is not?

Reference sheet

The explanations for all ten questions sit on this sheet. Pick an answer on the left and the matching card lights up.

T2A01 Explanation 1

Key A — 50 volts

E = I x R = 0.5 x 100 = 50 V. The 200 V answer comes from dividing instead of multiplying, which is the usual slip in this question.

Rule Fundamentals - Ohm's law

T2A02 Explanation 2

Key D — 24 watts

P = I x E = 2 x 12 = 24 W. Note that this is the power drawn from the supply, not the RF power delivered to the antenna - the difference is lost as heat in the transmitter.

Rule Fundamentals - power law

T2A03 Explanation 3

Key C — 24 ohms

R = E / I = 12 / 0.5 = 24 ohms. Multiplying instead gives 6, which is the trap answer.

Rule Fundamentals - Ohm's law rearranged

T2A04 Explanation 4

Key B — About 1.41 amperes

Use P = I^2 R, so I = square root of (P / R) = square root of (100 / 50) = square root of 2 = 1.41 A. The 2 A answer would be right for a 25 ohm load, and would also be the supply current if the voltage were 50 V.

Rule Fundamentals - power law in the form P = I^2 R

T2A05 Explanation 5

Key A — The ohm

The ohm is the resistance that allows one ampere to flow when one volt is applied. The volt measures potential difference, the ampere measures current, and the coulomb measures charge.

Rule Fundamentals - units

T2A06 Explanation 6

Key D — 600 ohms

Series resistances add directly: 100 + 200 + 300 = 600 ohms. The 55 ohm answer is the parallel combination, which is always smaller than the smallest individual resistor.

Rule Fundamentals - series circuits

T2A07 Explanation 7

Key C — 50 ohms

Two equal resistors in parallel give half the value, so 50 ohms. In parallel, current has two paths, so the combination must be lower than either resistor on its own.

Rule Fundamentals - parallel circuits

T2A08 Explanation 8

Key B — 6 watts

P = E^2 / R = 144 / 24 = 6 W. Equivalently, I = 12 / 24 = 0.5 A, and P = I x E = 0.5 x 12 = 6 W. Multiplying voltage by resistance instead gives the meaningless 288.

Rule Fundamentals - power dissipated in a resistance

T2A09 Explanation 9

Key A — 3 amperes

The lead resistance is in series with the load, so the total is 4 ohms and I = 12 / 4 = 3 A. This is why thick, short power leads matter: the current you thought you had is shared with the wire.

Rule Fundamentals - series circuits and voltage drop

T2A10 Explanation 10

Key A — P = E^2 / R

Substituting I = E / R into P = I x E gives P = E^2 / R. The form E / R gives current, not power, and E x R has no meaning at all.

Rule Fundamentals - power law variants

How to use this unit

Work the ten questions in one sitting with the reference sheet covered, then uncover it and read every explanation - including the ones you got right. The wording of the real exam is different, so learn the reasoning rather than the answer letters.