State and apply Ohm's Law (V = IR): Worked examples

Before we start, one reminder of what each letter means (never use a symbol you can't say in words):

$V$ = potential difference (the "push"), measured in volts (V) — see Potential Difference (Voltage).
$I$ = current (the "flow rate"), measured in amperes (A) — see Electric Current.
$R$ = resistance (the "clogging"), measured in ohms (Ω) — see Resistance.

The scenario matrix

Every Ohm's-Law question falls into one of these case classes. Each row is a different shape of problem; the last column tells you which worked example below hits it.

#	Case class	What's unknown / what's weird	Example
A	Solve for $I$	Given $V, R$ (base units)	Ex 1
B	Solve for $R$	Given $V, I$ (current < 1 A)	Ex 2
C	Solve for $V$	Given $I, R$ with prefix units (mA, kΩ)	Ex 3
D	Zero / degenerate input	$V = 0$ , or $R \to \infty$ (open), or $R \to 0$ (short)	Ex 4
E	Proportional reasoning	No numbers plugged — scale $V$ , watch $I$	Ex 5
F	Word problem (real world)	Translate English → $V, I, R$	Ex 6
G	Graph / geometric	Read $R$ off a $V$ – $I$ line's gradient	Ex 7
H	Non-ohmic twist	$R$ changes; find it at two points	Ex 8
I	Exam twist (multi-step)	Two resistors, must combine first	Ex 9

We work each cell in order. The method never changes:

(1) Write down what you have and what you want. (2) Pick the right form from the VIR triangle ( $V = I R$ , $I = V / R$ , $R = V / I$ ). (3) Convert to base units first, then compute, then check.

Case B — Solve for resistance (small current)

Case C — Solve for voltage (prefix units)

Case D — Zero and degenerate inputs

This is the row most notes skip. Every limiting case must have an answer.

Worked example Ex 4 — Three degenerate scenarios

(a) No push: $V = 0 V$ across a $50 Ω$ resistor. Find $I$ . (b) Open circuit: the wire is broken — treat the gap as $R \to \infty$ . What is $I$ when $V = 9 V$ ? (c) Short circuit: an ideal wire has $R \to 0$ . What does $I = V / R$ predict?

Forecast: (a) No push → no flow. (b) Infinite clog → no flow. (c) Zero clog → huge flow. Predict the numbers.

Step 1 — Case (a), $V = 0$ . $I = \frac{V}{R} = \frac{0}{50} = 0 A$ Why? Zero on top of any nonzero number is zero — no push, no current. This is the baseline everything else grows from.

Step 2 — Case (b), open circuit, $R \to \infty$ . $I = \frac{V}{R} = \frac{9}{very large} \to 0 A$ Why? Dividing a fixed 9 by an ever-larger clog drives the flow toward zero. A broken circuit carries no current — matches everyday life (switch off = gap = no flow).

Step 3 — Case (c), short circuit, $R \to 0$ . $I = \frac{V}{R} = \frac{9}{tiny} \to \infty$ Why? Dividing 9 by an ever-smaller clog makes flow blow up. In reality no wire is truly $0 Ω$ , so the current is huge-but-finite — this is why a short blows the fuse.

Verify: All three obey $V = I R$ . (a) $0 = 0 \times 50$ ✓. (b) as $R \to \infty$ , $I R = (9/ R) \cdot R = 9$ still ✓ — the product stays honest even at the limit. (c) as $R \to 0$ , current unbounded — physically capped by real wire resistance and the fuse.

Case E — Proportional reasoning (no plugging)

Case F — Real-world word problem

Worked example Ex 6 — The phone charger

A phone charger delivers a steady $5 V$ . Inside a test load, a current of $2 A$ is measured. Treating the load as an ohmic resistor, find its resistance — and then, using power, the power it dissipates.

Forecast: 5 V pushing a healthy 2 A means a low resistance (a few ohms). And $5 \times 2$ watts of heat.

Step 1 — Translate English → symbols. "delivers 5 V" ⇒ $V = 5 V$ . "current of 2 A" ⇒ $I = 2 A$ . Want $R$ . Why this step? Word problems are 90% translation; once labelled, it's just Ex 2 again.

Step 2 — Resistance. Want $R$ → $R = V / I$ . $R = \frac{5}{2} = 2.5 Ω$

Step 3 — Power (bonus). $P = V I = 5 \times 2 = 10 W$ . Why this step? Real chargers are rated in watts; this links Ohm's Law to what the label says.

Verify: $V = I R = 2 \times 2.5 = 5 V$ ✓. $P = V I = 10 W$ ; cross-check $P = I^{2} R = 4 \times 2.5 = 10 W$ ✓.

Case G — Graph / geometric (gradient = resistance)

Worked example Ex 7 — Read resistance off the V–I line

The graph below plots $V$ (vertical) against $I$ (horizontal) for an ohmic resistor. The straight line passes through the origin and through the point $(I, V) = (0.4 A, 6 V)$ . Find $R$ .

Forecast: For $V$ -vs- $I$ , the gradient (rise over run) equals $R$ , because $V = I R$ rearranges to $slope = V / I = R$ . Steeper line ⇒ bigger resistance. Eyeball it: is this steep or shallow?

Step 1 — Identify the tool: gradient. Why gradient and not just any point? Because for a straight line through the origin, every point gives the same ratio $V / I$ — that constant ratio is exactly what "resistance" means (see Resistance).

Step 2 — Compute rise over run. Look at the red triangle in the figure: rise $= 6 V$ , run $= 0.4 A$ . $R = gradient = \frac{Δ V}{Δ I} = \frac{6}{0.4} = 15 Ω$

Step 3 — Cross-check with a second point. The pale-blue dashed line to $(0.2 A, 3 V)$ gives $3/0.2 = 15 Ω$ — same value, proving the line is genuinely ohmic (constant $R$ ).

Verify: $V = I R = 0.4 \times 15 = 6 V$ ✓ and $0.2 \times 15 = 3 V$ ✓. Both points sit on the line.

Case H — Non-ohmic twist ( $R$ changes)

Worked example Ex 8 — Filament lamp: resistance at two brightnesses

A filament lamp is measured twice.

Cold/dim: $2 V$ gives $1 A$ .
Hot/bright: $8 V$ gives $2 A$ . Find $R$ at each point. Is it ohmic?

Forecast: Voltage went ×4 but current only ×2 — the flow did not keep up with the push. That means the clogging grew. Expect $R$ to rise.

Step 1 — Resistance when dim. $R_{1} = V / I = 2/1 = 2 Ω$ . Why this step? $R = V / I$ still defines an instantaneous resistance even for non-ohmic parts.

Step 2 — Resistance when bright. $R_{2} = V / I = 8/2 = 4 Ω$ .

Step 3 — Compare. $R_{2} = 4 Ω > R_{1} = 2 Ω$ . The resistance doubled. Why? The hot filament's atoms vibrate more → more electron collisions → higher resistance. Not constant ⇒ not ohmic ⇒ its $V$ – $I$ graph curves.

Verify: If it were ohmic, ×4 voltage would give ×4 current (8 A). It gave only 2 A, so the ratio $V / I$ changed from 2 to 4 — proof of non-ohmic behaviour. Both points check: $2 = 1 \times 2$ ✓, $8 = 2 \times 4$ ✓.

Case I — Exam twist (combine first)

Worked example Ex 9 — Two resistors in series, then Ohm's Law

A $9 V$ battery drives two resistors in series: $R_{1} = 2 Ω$ and $R_{2} = 4 Ω$ . Find the current from the battery, then the voltage across $R_{2}$ .

Forecast: Series resistors add up, so the battery "sees" a bigger clog than either alone. Predict a current below $9/4 = 2.25 A$ .

Step 1 — Combine the resistance. In series resistances add: $R_{total} = R_{1} + R_{2} = 2 + 4 = 6 Ω$ . Why this step? Ohm's Law needs one resistance across one voltage; you must reduce the network first.

Step 2 — Total current. Same current flows through everything in series: $I = \frac{V}{R _{total}} = \frac{9}{6} = 1.5 A$

Step 3 — Voltage across $R_{2}$ . Apply Ohm's Law to that resistor alone, using its own current (which equals the series current $1.5 A$ ): $V_{2} = I R_{2} = 1.5 \times 4 = 6 V$ Why this step? "Voltage is across a component, current is through it" — use $R_{2}$ 's own share.

Verify: $V_{1} = I R_{1} = 1.5 \times 2 = 3 V$ . Then $V_{1} + V_{2} = 3 + 6 = 9 V$ = battery voltage ✓ (voltages in series must add back to the source). And $1.5 A < 2.25 A$ as forecast.

Active Recall

Recall Q: A broken (open) circuit has

$R \to \infty$ . What current flows? $I = V / R \to 0 A$ — no current. A gap stops the flow.

Recall Q: On a

$V$ -vs- $I$ graph, a line through $(0.4 A, 6 V)$ has what resistance? Gradient $= 6/0.4 = 15 Ω$ .

Recall Q: A lamp reads 2 Ω dim and 4 Ω bright. Ohmic?

No — $R$ changed, so it is non-ohmic; its $V$ – $I$ graph curves.

Recall Q: 30 mA through 2.2 kΩ — voltage? (use the shortcut)

mA × kΩ = V, so $30 \times 2.2 = 66 V$ .

Recall Q: Series 2 Ω + 4 Ω on 9 V — battery current?

$R_{total} = 6 Ω$ , $I = 9/6 = 1.5 A$ .

80/20 — the scenario habits

Name Have/Want first, then pick the triangle form.
Convert prefixes (mA, kΩ) before plugging.
Limits matter: $V = 0 \Rightarrow I = 0$ ; open $\Rightarrow I \to 0$ ; short $\Rightarrow I \to \infty$ .
Graphs: gradient of $V$ -vs- $I$ = $R$ ; curved = non-ohmic.
Networks: combine resistances, then apply $V = I R$ ; series voltages add back to the source.

Connections

↑ Parent: Ohm's Law (V = IR)
Electric Current · Potential Difference (Voltage) · Resistance
Series and Parallel Circuits — needed for Ex 9
Power in Circuits (P = VI) — used in Ex 6
Non-ohmic Components — the twist in Ex 8

Worked examples — State and apply Ohm's Law (V = IR)