Define resistance and the ohm: Worked examples — Hardware deep dive

We build only on the one equation and its two rearrangements. Everything is that triangle:

The scenario matrix

Before working anything, let's list every distinct kind of case this topic can throw at you. Each worked example below is tagged with the cell it fills.

#	Case class	What makes it tricky	Example
C1	Find $R$ — clean values	nothing; the base skill	Ex 1
C2	Find $I$ — rearrange	must solve for the denominator	Ex 2
C3	Find $V$ — rearrange	must multiply back	Ex 3
C4	Unit prefixes (µA, mA, kΩ, mV)	hidden powers of ten	Ex 4
C5	Zero input ( $I = 0$ , $V = 0$ , or both)	division by zero / open, short & indeterminate	Ex 5
C6	Limiting behaviour ( $R \to 0$ , $R \to \infty$ )	what happens at the extremes	Ex 6
C7	Graph reading (I–V slope)	slope is $1/ R$ , not $R$	Ex 7
C8	Real-world word problem	strip the story to $V, I, R$	Ex 8
C9	Exam twist (change one quantity)	proportional reasoning, no numbers	Ex 9
C10	Non-ohmic sanity check	$R = V / I$ still works pointwise	Ex 10
C11	Negative sign convention ( $V < 0$ or $I < 0$ )	direction, not smaller-than-nothing	Ex 11

We now clear each cell.

C1 — Find $R$ from clean values

C2 — Rearrange to find current

C3 — Rearrange to find voltage

C4 — Unit prefixes (the powers-of-ten trap)

Before we compute, we must earn the prefixes. They are just shorthand for powers of ten:

Worked example Example 4 (cell C4)

(a) A resistor carries $5 mA$ when $2 V$ is placed across it. Find $R$ in kilohms. (b) A sensor lets only $50 μ A$ through when $10 V$ is applied. Find $R$ in megohms.

Forecast: Tiny currents (milliamps, then microamps) with a few volts. Big or small resistance — and which part gives the bigger $R$ ?

Step 1 (part a). Convert to base units first: $5 mA = 5 \times 1 0^{- 3} A = 0.005 A$ Why this step? The law $R = V / I$ only balances when $V$ is in volts and $I$ is in amperes. Mixing a milliamp in raw would make the answer 1000× wrong.

Step 2 (part a). Apply the definition: $R = \frac{V}{I} = \frac{2 V}{0.005 A} = 400 Ω = 0.4 k Ω$ Why this step? Base units in, base unit out (ohms), then divide by 1000 for kilohms.

Step 3 (part b). Convert the microamp current: $50 μ A = 50 \times 1 0^{- 6} A = 0.000050 A$ Why this step? A micro is a millionth, so this is even smaller than a milliamp — the factor is $1 0^{- 6}$ , not $1 0^{- 3}$ .

Step 4 (part b). Apply the definition and convert: $R = \frac{10 V}{0.000050 A} = 200, 000 Ω = 0.2 M Ω$ Why this step? An even tinier current for a bigger voltage means an even bigger resistance; $1 M Ω = 1 0^{6} Ω$ , so divide by a million for megohms.

Verify: Part (a): $I = V / R = 2/400 = 0.005 A = 5 mA$ ✔. Part (b): $I = 10/200000 = 0.000050 A = 50 μ A$ ✔. The microamp case gave the far larger resistance, as forecast.

C5 — Zero inputs (open circuit, short circuit, and the 0/0 origin)

Zeros break naive division, so we must reason about them physically, not just plug in.

Worked example Example 5 (cell C5)

Three puzzle parts: (a) A gap in the wire (a broken switch) means no current flows, $I = 0$ , even though $V = 12 V$ sits across the gap. What is the "resistance" of the gap? (b) A perfect wire (a short) has essentially no voltage across it, $V = 0$ , while a current $I = 3 A$ flows. What is its resistance? (c) A resistor sits with the supply switched off: $V = 0$ and $I = 0$ at the same time. What does $R = V / I$ give, and does the resistor still have a resistance?

Forecast: Part (a) gives a giant resistance, part (b) a near-zero one. But part (c) is the sneaky one — what can $0/0$ possibly mean?

Step 1 (part a). Try $R = V / I = 12/0$ . Why this step? To see what the equation says — and it says "divide by zero", which is undefined and blows up.

Step 2 (part a). Interpret physically: a value that grows without limit as $I \to 0$ means the resistance is infinitely large — an open circuit. No path, no current, no matter how hard you push. $R \to \infty (open circuit)$ Why this step? We swap the meaningless "12/0" for its limiting meaning: infinite opposition.

Step 3 (part b). Apply the definition: $R = \frac{V}{I} = \frac{0 V}{3 A} = 0 Ω$ Why this step? Zero push for real flow means zero opposition — a short circuit / ideal conductor. See Conductors and Insulators.

Step 4 (part c). Try the switched-off resistor: $R = \frac{V}{I} = \frac{0}{0} = indeterminate$ Why this step? $0/0$ is not a number — the formula simply cannot report a resistance when nothing is being pushed and nothing is flowing. The origin $(0, 0)$ lies on every I–V line through zero, so a single point at the origin cannot pick out which resistance you have.

Step 5 (part c — the fix). Resistance is a property of the material, not something the origin measures. To read it, you must apply some non-zero voltage and see the resulting current, i.e. use any other point on the line (or its slope). The resistor still has, say, $4 Ω$ — you just can't extract it from the dead point $(0, 0)$ . Why this step? It separates "the formula is undefined here" from "the physical resistance is undefined" — only the first is true.

Verify: Part (a): $I = V / R = 12/\infty \to 0 A$ ✔ (no current, as stated). Part (b): $V = I R = 3 \times 0 = 0 V$ ✔ (no voltage drop, as stated). Part (c): a $4 Ω$ resistor at $V = 0$ indeed gives $I = 0/4 = 0 A$ , so $(0, 0)$ is consistent with $4 Ω$ — and equally with $9 Ω$ ( $0/9 = 0$ ) — confirming the origin alone cannot fix $R$ ✔.

C6 — Limiting behaviour

Figure s01 — what you are looking at. The horizontal axis is resistance $R$ in ohms (from $1$ to $30 Ω$ ); the vertical axis is current $I$ in amperes. The amber curve is $I = V / R$ with the voltage held fixed at $12 V$ — so as $R$ changes, only the current is allowed to respond. The cyan dot marks the anchor point $(12 Ω, 1 A)$ . Notice the shape: on the left the curve rockets almost straight up (the " $R \to 0$ " arrow) — dividing $12$ by a tiny number gives a huge current. On the right it flattens toward the horizontal axis (the " $R \to \infty$ " arrow) — dividing $12$ by a huge number gives almost nothing. This bending-down curve is a hyperbola, the visual signature of an inverse relationship: double $R$ and the current halves.

Worked example Example 6 (cell C6)

With $V = 12 V$ held constant, describe what happens to the current as $R \to 0$ and as $R \to \infty$ .

Forecast: Which end of the amber curve gives a flood of current, which end a trickle?

Step 1. Use $I = V / R$ and shrink $R$ toward zero: $R \to 0 ⟹ I = \frac{12}{R} \to \infty$ Why this step? Dividing a fixed number by an ever-smaller number gives an ever-larger result — look at the curve rocketing up on the left of the figure.

Step 2. Now grow $R$ toward infinity: $R \to \infty ⟹ I = \frac{12}{R} \to 0$ Why this step? Dividing a fixed number by an ever-bigger number gives an ever-smaller result — the curve flattens onto the axis on the right.

Step 3 (the anchor point on the curve). At $R = 12 Ω$ : $I = \frac{12}{12} = 1 A$ Why this step? One labelled point (the cyan dot) anchors the whole curve — it passes through $(12 Ω, 1 A)$ .

Verify: At $R = 6 Ω$ , $I = 12/6 = 2 A$ (higher, to the left) ✔; at $R = 24 Ω$ , $I = 12/24 = 0.5 A$ (lower, to the right) ✔. The curve is a decreasing hyperbola exactly as drawn.

C7 — Reading a graph (slope is $1/ R$ , not $R$ )

Figure s02 — what you are looking at. This is an I–V characteristic: the horizontal axis is voltage $V$ in volts and the vertical axis is current $I$ in amperes. The amber line is straight and passes through the origin — the fingerprint of an ohmic resistor obeying Ohm's Law. The cyan dot marks the reading $(8 V, 2 A)$ . The white dashed right-angle triangle underneath shows how to measure the slope: the horizontal "run" is $8 - 4 = 4 V$ and the vertical "rise" is $2 - 1 = 1 A$ , so the slope is $\frac{1}{4}$ . The amber caption reminds you that this slope equals $1/ R$ , not $R$ — so a steeper line would mean a smaller resistance. Read your gaze from origin, up the line, to the dot, then down the dashed triangle to see rise-over-run.

Worked example Example 7 (cell C7)

From the graph, the line passes through the origin and through the point $(V, I) = (8 V, 2 A)$ . Find the resistance.

Forecast: The slope of this line is $2/8 = 0.25$ . Is the resistance $0.25 Ω$ , or something else?

Step 1. Define slope on an I–V graph: $slope = \frac{rise}{run} = \frac{I}{V} = \frac{2 A}{8 V} = 0.25 A/V$ Why this step? Slope is always vertical-change over horizontal-change; here that ratio is $I / V$ .

Step 2. Relate slope to resistance. Since $R = V / I$ , its reciprocal is: $\frac{1}{R} = \frac{I}{V} = slope$ Why this step? The graph's slope is $I / V$ , but $R$ is the other ratio $V / I$ — they are reciprocals. So resistance is $1/ slope$ , never the slope itself.

Step 3. Take the reciprocal: $R = \frac{1}{0.25 A/V} = 4 Ω$ Why this step? Flip the slope to recover $R$ in ohms.

Verify: Read one point directly: $R = V / I = 8/2 = 4 Ω$ ✔. A steeper line (bigger slope) would mean a smaller $R$ — see Ohm's Law for why the straight line means "ohmic".

C8 — Real-world word problem

C9 — Exam twist (proportional reasoning, no fresh numbers)

Worked example Example 9 (cell C9)

A resistor obeys $V = I R$ . You triple the voltage across it while the resistance stays the same. What happens to the current? Then, separately, you triple the resistance at fixed voltage — now what happens to the current?

Forecast: Predict both before doing the algebra: does tripling voltage help or hurt the current? Does tripling resistance help or hurt it?

Step 1. Write current as a fraction so the two variables are visible: $I = \frac{V}{R}$ Why this step? To see how $I$ responds, we need it isolated with $V$ on top and $R$ on the bottom.

Step 2 — triple $V$ (keep $R$ ). Replace $V$ by $3 V$ : $I_{new} = \frac{3 V}{R} = 3 \cdot \frac{V}{R} = 3 I$ Why this step? $V$ sits in the numerator, so multiplying it by 3 multiplies the whole current by 3 — current is directly proportional to voltage.

Step 3 — triple $R$ (keep $V$ ). Replace $R$ by $3 R$ : $I_{new} = \frac{V}{3 R} = \frac{1}{3} \cdot \frac{V}{R} = \frac{I}{3}$ Why this step? $R$ sits in the denominator, so multiplying it by 3 divides the current by 3 — current is inversely proportional to resistance.

Verify (with sample numbers). Take $V = 12 V$ , $R = 4 Ω$ , so $I = 3 A$ . Triple $V$ : $36/4 = 9 A = 3 \times 3$ ✔. Triple $R$ : $12/12 = 1 A = 3/3$ ✔.

C10 — Non-ohmic sanity check

Not every component gives a straight I–V line. A filament lamp curves because it heats up and its resistance rises. But $R = V / I$ still works — at each single point.

Worked example Example 10 (cell C10)

A filament lamp measures $(V_{1}, I_{1}) = (2 V, 0.5 A)$ at low brightness and $(V_{2}, I_{2}) = (8 V, 1 A)$ at high brightness. Find the resistance at each point and comment.

Forecast: For an ohmic resistor $V / I$ would be identical at both points. Do you expect these two to match?

Step 1. Compute $R$ at the first point: $R_{1} = \frac{V _{1}}{I _{1}} = \frac{2 V}{0.5 A} = 4 Ω$ Why this step? $R = V / I$ is defined at any single operating point, even when the curve isn't straight.

Step 2. Compute $R$ at the second point: $R_{2} = \frac{V _{2}}{I _{2}} = \frac{8 V}{1 A} = 8 Ω$ Why this step? Same definition, different point — because the filament is hotter, its resistance is larger.

Step 3. Compare the two and conclude: $R_{2} = 8 Ω > R_{1} = 4 Ω$ Why this step? A rising $R$ with brightness is the fingerprint of a non-ohmic device — the I–V line would bend upward-then-flatten, not stay straight. So $R = V / I$ is still valid, but it gives a different number at each point rather than one fixed resistance.

Verify: For an ohmic part $R_{1}$ would equal $R_{2}$ . Here they differ ( $4 Ω$ vs $8 Ω$ ), correctly flagging non-ohmic behaviour. Push back: $V_{2} = I_{2} R_{2} = 1 \times 8 = 8 V$ ✔ and $V_{1} = I_{1} R_{1} = 0.5 \times 4 = 2 V$ ✔.

C11 — Negative signs (direction, not "less than nothing")

A minus sign on $V$ or $I$ does not mean a negative resistance. It means the direction reversed — you swapped which terminal you called "plus", or the current runs the other way round the loop. Resistance itself is always a positive property of the material.

Worked example Example 11 (cell C11)

A resistor is measured with the meter leads reversed, so the readings come out as $V = - 6 V$ and $I = - 2 A$ . Find the resistance. Then a second reading on the same resistor comes out as $V = - 6 V$ while $I = + 2 A$ — what does that tell you?

Forecast: Will the reversed-lead case give a negative resistance, or the same positive value as if the leads were the right way round?

Step 1 (reversed leads). Apply the definition with both signs kept: $R = \frac{V}{I} = \frac{- 6 V}{- 2 A} = + 3 Ω$ Why this step? A negative divided by a negative is positive — the two minus signs cancel. Flipping the leads flips both readings together, so their ratio is unchanged.

Step 2 (interpret). The resistance is $+ 3 Ω$ , exactly as if the leads were the normal way round. Why this step? Resistance is a positive property; the signs only recorded that we measured "backwards", not that the path fights the flow any differently.

Step 3 (the mismatched-sign case). Now $V$ and $I$ have opposite signs: $R = \frac{- 6 V}{+ 2 A} = - 3 Ω$ Why this step? A genuine ohmic resistor can never produce this — a real material always pushes current the same way the voltage points, so $V$ and $I$ must share a sign. A computed $R < 0$ is therefore a red flag: you have almost certainly mislabelled which reading points which way, or you are not looking at a plain passive resistor at all.

Verify: Reversed-lead ratio: $(- 6) / (- 2) = 3 Ω > 0$ ✔ — same magnitude as $6/2$ . Sanity rule: for any passive resistor, $V$ and $I$ always carry the same sign, so $R = V / I$ is always positive; a negative value means a labelling error, not a real negative resistance.

Recall

Recall Every cell, one line each

Cell C1 — find R from clean values ::: $R = V / I$ directly (Ex 1: $6/2 = 3 Ω$ ). Cell C2 — find current ::: $I = V / R$ (Ex 2: $9/45 = 0.2 A$ ). Cell C3 — find voltage ::: $V = I R$ (Ex 3: $0.4 \times 25 = 10 V$ ). Cell C4 — convert prefixes first ::: $5 mA = 0.005 A$ so $R = 400 Ω = 0.4 k Ω$ ; $50 μ A = 0.000050 A$ so $R = 200000 Ω = 0.2 M Ω$ . Cell C5 — $I = 0$ means ::: open circuit, $R \to \infty$ . Cell C5 — $V = 0$ with current means ::: short circuit, $R = 0 Ω$ . Cell C5 — $V = 0$ and $I = 0$ together ::: $0/0$ is indeterminate; the origin cannot fix $R$ , you must use another point. Cell C6 — $R \to 0$ at fixed $V$ ::: current $\to \infty$ . Cell C6 — $R \to \infty$ at fixed $V$ ::: current $\to 0$ . Cell C7 — I–V graph slope equals ::: $1/ R$ , so $R = 1/ slope$ . Cell C9 — triple $V$ at fixed $R$ ::: current triples (direct proportion). Cell C9 — triple $R$ at fixed $V$ ::: current thirds (inverse proportion). Cell C10 — non-ohmic component ::: $V / I$ differs between points, so $R$ changes. Cell C11 — both $V$ and $I$ negative ::: signs cancel, $R = V / I > 0$ (direction, not negative resistance).

Connections

Define resistance and the ohm — the parent note these examples drill.
Ohm's Law — why the I–V line is straight for ohmic parts (and bends for lamps).
Voltage and Potential Difference — the $V$ in every example, including its sign.
Electric Current — the $I$ in every example, and the µA/mA conversions.
Conductors and Insulators — the $R = 0$ and $R \to \infty$ extremes of Ex 5.
Resistivity and Resistance of a Wire — what physically sets $R$ before you divide.
Power Dissipation in Resistors — why the low-resistance kettle element of Ex 8 gets hot.