Calculate electrical power (P = VI, P = I²R): Worked examples

Before anything else, let us be crystal-clear on the three letters, because every example below uses them.

The scenario matrix

Electrical power problems don't have "negative angles" or "quadrants" like trig — but they DO have distinct case classes. Here is every cell this topic can throw at you:

#	Case class	What you're given	Right tool	Edge/twist
A	Base case	$V$ and $I$	$P = V I$	none
B	No voltage known	$I$ and $R$	$P = I^{2} R$	leads to power rating
C	No current known	$V$ and $R$	$P = V^{2} / R$	none
D	Energy over time	$P$ and $t$	$E = P t$	unit conversion (h→s)
E	Zero / degenerate	$I = 0$ , or $R \to \infty$ , or $R = 0$	any	limiting behaviour
F	Scaling / proportion	"double the current"	$P \propto I^{2}$	the ×4 trap
G	Real-world word problem	mixed, must extract data	pick from trio	battery / bill
H	Exam twist — series divider	supply $V$ , one resistor	$P = I^{2} R$ per part	wrong- $V$ trap

The eight examples below hit every cell A–H. Each is labelled with its cell.

Worked examples

The next example is the one where a picture earns its keep, because "what happens as we push resistance to extremes" is best seen as a curve.

Worked example Example 5 — Cell E (zero / degenerate / limiting cases)

A fixed $12 V$ supply sits across a resistor $R$ . Work out the power for three extreme values of $R$ : (a) $R = 0$ (a dead short), (b) $R = 6 Ω$ (normal), (c) $R \to \infty$ (open circuit / broken wire).

Forecast: Which case gives the MOST power? Which gives ZERO? Guess before the maths.

Use $P = V^{2} / R$ because voltage is fixed and we're varying $R$ . Why this step? With $V$ held constant, this form shows the dependence on $R$ cleanly — it's power vs one variable.
(b) Normal, $R = 6 Ω$ : $P = \frac{1 2 ^{2}}{6} = \frac{144}{6} = 24 W$
(c) Open circuit, $R \to \infty$ : dividing $144$ by a huge number → $P \to 0 W$ . Why this step? No path for charge means $I = 0$ ; with no current, no energy is delivered. Look at the magenta curve in the figure — it flattens toward zero on the right.
(a) Dead short, $R = 0$ : algebraically $P = 144/0 \to \infty$ . Why this step? This is the danger limit — a real supply has internal resistance, but the formula screams "unbounded heat," which is exactly why a short circuit melts wires and pops fuses. On the figure this is the curve rocketing up on the left (the orange dashed asymptote).

Verify: Sanity ladder — as $R$ grows, $P = 144/ R$ must shrink: $R = 6 \to 24$ W, $R = 12 \to 12$ W, $R = 144 \to 1$ W. Monotonic decrease ✓. And $I = V / R$ at $R \to \infty$ gives $I = 0$ , confirming $P = V I = 0$ ✓. See Heat Dissipation and Cooling for why the short case is lethal.

Worked example Example 6 — Cell F (scaling / the ×4 trap)

A heater element has fixed resistance $R = 10 Ω$ . At current $I_{1} = 2 A$ it dissipates some power $P_{1}$ . We now double the current to $I_{2} = 4 A$ . By what factor does the power grow?

Forecast: Most people say "double." Write down your gut answer first.

Compute $P_{1} = I_{1}^{2} R = 2^{2} \times 10 = 40 W$ . Why this step? $R$ is fixed and we know $I$ , so $P = I^{2} R$ is the natural tool — and it exposes the square.
Compute $P_{2} = I_{2}^{2} R = 4^{2} \times 10 = 160 W$ .
Ratio: $\frac{P _{2}}{P _{1}} = \frac{160}{40} = 4$ Why this step? Because $P \propto I^{2}$ , doubling $I$ multiplies power by $2^{2} = 4$ , not 2. In the figure the violet bars double in $I$ but quadruple in height.

Verify: Symbolically, $\frac{P _{2}}{P _{1}} = \frac{I _{2}^{2} R}{I _{1}^{2} R} = (\frac{I _{2}}{I _{1}})^{2} = 2^{2} = 4$ ✓. This is why doubling current in a wire produces four times the heat — the core reason thick, low-current wiring is used in power delivery.

Worked example Example 7 — Cell G (real-world word problem)

A phone battery is labelled $3.7 V$ , $4000 mAh$ (milliamp-hours). While gaming, the phone draws $I = 1.5 A$ . (i) What power is the battery delivering? (ii) Roughly how many hours of gaming does a full charge give?

Forecast: Guess the power (a few watts?) and the runtime (more or less than 3 hours?).

Extract the data: $V = 3.7 V$ , $I = 1.5 A$ , capacity $= 4000 mAh = 4 Ah$ . Why this step? Word problems hide numbers in prose — list them with units first so you don't mix $mAh$ and $Ah$ .
(i) Power delivered: we have $V$ and $I$ , so $P = V I = 3.7 \times 1.5 = 5.55 W$ Why this step? Cell A tool — both quantities are given directly.
(ii) Runtime: capacity in amp-hours ÷ draw in amps gives hours. $t = \frac{4 Ah}{1.5 A} \approx 2.67 h$ Why this step? $Ah / A = h$ : the amps cancel, leaving hours of supply.

Verify: Energy check: $E = P \cdot t = 5.55 W \times (2.67 \times 3600 s) \approx 53300 J$ . Independently, battery energy $\approx V \times charge = 3.7 \times (4 \times 3600 C) = 53280 J$ ✓. The two agree, so the runtime is consistent.

Worked example Example 8 — Cell H (exam twist: the series-divider trap)

A $12 V$ supply drives two resistors in series: $R_{1} = 4 Ω$ and $R_{2} = 8 Ω$ . Find the power dissipated in $R_{1}$ alone.

Forecast: The trap answer is $P = V^{2} / R_{1} = 1 2^{2} /4 = 36 W$ . Is that right? Guess yes/no.

Find the total resistance: in series they add, $R_{tot} = 4 + 8 = 12 Ω$ . Why this step? The same current flows through both, and it's set by the whole chain.
Find the shared current using Ohm's law on the whole loop: $I = \frac{V}{R _{tot}} = \frac{12}{12} = 1 A$ Why this step? In series there is one current everywhere; the supply voltage divides across the resistors, so $12 V$ is NOT the voltage across $R_{1}$ by itself.
Power in $R_{1}$ using $P = I^{2} R$ with that current: $P_{1} = I^{2} R_{1} = 1^{2} \times 4 = 4 W$ Why this step? $I^{2} R$ is per-resistor safe because it uses the actual current and that resistor's own $R$ — no wrong voltage to grab.

Verify: Cross-check with the correct voltage across $R_{1}$ : $V_{1} = I R_{1} = 1 \times 4 = 4 V$ , so $P_{1} = V_{1}^{2} / R_{1} = 16/4 = 4 W$ ✓. Also total power $P = V I = 12 \times 1 = 12 W$ , and $P_{1} + P_{2} = 4 + (1^{2} \times 8) = 4 + 8 = 12 W$ ✓ — the parts add up to the whole. The naive $36 W$ used the whole-circuit voltage on one resistor — that's the trap.

Recall Quick self-test (reveal after guessing)

Which formula when you know $I$ and $R$ but not $V$ ? ::: $P = I^{2} R$ A fixed $V$ across resistance $R \to \infty$ gives what power? ::: Zero — no current flows. Double the current in a fixed resistor: power ×? ::: ×4, because $P \propto I^{2}$ . $5 V \times 3 A$ charger delivers? ::: $15 W$ Series $R_{1} = 4 Ω, R_{2} = 8 Ω$ on $12 V$ : power in $R_{1}$ ? ::: $4 W$ (not 36 W) Convert $3$ hours to seconds for a joule calc. ::: $10800 s$

Connections

Ohm's Law (V = IR) — the bridge used in Cells B, C, E, H.
Voltage and Potential Difference — the $V$ that must be the right $V$ (Cell H).
Electric Current — the shared series current (Cell H) and the $I = 0$ limit (Cell E).
Energy and the Joule — $E = P t$ and the battery-energy check (Cells D, G).
Resistors and Power Ratings — the $0.25 W$ survival check (Cell B).
Heat Dissipation and Cooling — why the short-circuit limit (Cell E) is dangerous.