State and apply Ohm's Law (V = IR): Question bank — Hardware deep dive

Before we start, four ideas we lean on constantly, in plain language. The last one (sign convention) is new here, so we build it with a picture.

Voltage ( $V$ ) — the push, measured across two points (energy handed to each unit of charge).
Current ( $I$ ) — the flow rate, measured through a point (charge passing per second).
Resistance ( $R$ ) — the opposition, defined as $R = V / I$ (how much push you needed per unit of flow you got).
Sign convention — a chosen positive direction for current and a chosen positive terminal for voltage. Reverse the battery and both $V$ and $I$ flip sign together; $R = V / I$ stays positive.

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

The figure above shows the full V–I graph. Look at the two straight lines through the origin: the magenta ohmic resistor runs into the third quadrant (bottom-left) too — negative voltage gives negative current, but the gradient $Δ V /Δ I = R$ is the same everywhere. The orange curve bends: that is a non-ohmic component whose slope changes. Refer back to this picture whenever a question below mentions signs, quadrants, or the graph's shape.

True or false — justify

True or false: If you double the voltage across an ohmic resistor, its resistance also doubles.

False.

R

is a fixed property of the material and geometry at constant temperature; doubling

V

doubles the current

I

, leaving the ratio

V / I = R

unchanged.

True or false: Ohm's Law

V = I R

works for absolutely every electrical component.

False. It holds only for ohmic conductors (constant

R

); for a filament lamp or diode,

R

changes with conditions, so the straight-line proportionality breaks — see Non-ohmic Components.

True or false: A component with a straight-line graph of

V

(y) against

I

(x) through the origin is ohmic.

True. A straight line through the origin — running into both the first and third quadrants — means

V / I

is the same for every point, i.e.

R

is constant, the definition of ohmic (magenta line in the figure).

True or false: For a fixed resistor, halving the current means the voltage across it halves.

True. With

R

constant,

V = I R

is directly proportional, so any change in

I

scales

V

by the same factor.

True or false: Resistance has no meaning for a non-ohmic device.

False.

R = V / I

still defines an instantaneous resistance at each point; it simply isn't a single fixed number — it changes as

V

changes (the orange curve's local slope varies).

True or false: If you reverse the battery, an ohmic resistor's resistance becomes negative.

False. Reversing the battery flips the signs of both

V

and

I

together, so

R = V / I

is still a positive number — the graph line just extends into the third quadrant with the same gradient.

True or false: A thicker wire of the same material and length has higher resistance.

False. A wider "pipe" lets charge flow more easily, so a thicker wire has lower resistance — see Resistance for the geometry.

True or false: For a passive resistor, current cannot exist without a voltage across it.

True as a relationship, not a cause-and-effect story:

I = V / R

, so zero across it means zero through it. But

V

and

I

arise together the instant the circuit closes — one does not happen "before" the other.

Spot the error

Spot the error: "A lamp draws 2 A at 4 V, so its resistance is fixed at 2 Ω for all voltages."

The mistake is fixed.

R = V / I = 2 Ω

only at that one operating point; a filament lamp heats up and its

R

rises at higher voltage, so it is non-ohmic.

Spot the error: "The resistor is

4.7 k Ω

and current is

2 mA

, so

V = 4.7 \times 2 = 9.4 V

The number is right, but only because the units quietly cancel — show them:

2 mA \times 4.7 k Ω = (2 \times 1 0^{- 3} A) \times (4.7 \times 1 0^{3} Ω) = 2 \times 4.7 \times 1 0^{- 3 + 3} V = 9.4 V

. The

1 0^{- 3}

from milli and

1 0^{3}

from kilo multiply to

1 0^{0} = 1

, which is exactly why "

mA \times k Ω = V

Spot the error: "To find the current in one resistor, use the total battery voltage in

I = V / R

You must use the voltage across that specific resistor, not the whole battery, unless that resistor is the only thing across the battery. In Series and Parallel Circuits the battery voltage splits between components.

Spot the error: "Current is measured across a component and voltage through it."

Reversed. Voltage is measured across (between two points, in parallel); current is measured through (along the path, in series).

Spot the error: "As a wire heats up, its resistance drops because heat gives electrons more energy."

For a metal, heating raises

R

: hotter atoms vibrate more, so electrons collide more often and flow is impeded. More thermal energy in the lattice hinders, not helps, the drift.

Spot the error: "The V–I graph is a straight line, so its gradient (with V on y, I on x) is the current."

The gradient is

\frac{Δ V}{Δ I} = \frac{V}{I} = R

, the resistance — not the current. Current is the x-axis value itself.

Spot the error: "The current came out negative, so I must have made an arithmetic mistake."

Not necessarily. A negative

I

just means the real flow is opposite to the arrow you chose — a legitimate sign-convention result, corresponding to the third quadrant of the V–I graph.

Spot the error: "

250 mA

through

220 Ω

gives

V = 250 \times 220 = 55000 V

Forgot to convert.

250 mA = 0.25 A

, so

V = 0.25 \times 220 = 55 V

— a factor of 1000 error from using milliamps directly.

Why questions

Why must temperature be held constant for Ohm's Law to hold?

Because

R

depends on temperature — heating a metal increases lattice vibration and electron collisions, raising

R

. If

R

drifts while you change

V

, the

V / I

ratio is no longer constant and proportionality fails.

Why is it misleading to say "voltage causes current"?

In a resistor

V

and

I

appear simultaneously and are locked by

V = I R

; neither precedes the other in time. It is safer to say the potential difference is the condition under which current flows — a relationship, not a one-way cause.

Why does the V–I graph of an ohmic resistor pass exactly through the origin and continue into the third quadrant?

V = 0

there is

I = 0

, so

(0, 0)

is on the line. Reversing the push makes both

V

and

I

negative while keeping

V / I = R

, so the same straight line extends symmetrically into the bottom-left (magenta line).

Why can a steeper V–I line (V on y-axis) mean a larger resistance?

The gradient equals

V / I = R

, so a steeper slope means more volts are needed per amp of current — exactly what "harder to push charge through" (bigger

R

) means.

Why is it wrong to say "Ohm's Law defines resistance"?

Resistance is defined by

R = V / I

for any component. Ohm's Law is the extra experimental claim that for certain materials this ratio stays constant — a discovery, not a definition.

Edge cases

Edge case: What is the current through a resistor when the voltage across it is exactly zero?

Zero.

I = V / R = 0/ R = 0

— no push, no flow, regardless of how small

R

is (as long as

R \neq = 0

Edge case: A resistor reads

I = 0

; does that force the voltage across that resistor to be zero?

Yes — for any ideal resistor,

V = I R = 0 \times R = 0

whenever

I = 0

, no matter where the break in the loop is. The subtlety is bookkeeping: if the loop is broken by an open switch, the battery voltage appears across the open switch (the gap), not across the resistor. So separate the two components — zero volts across the resistor, full battery volts across the open gap.

Edge case: What does

V = I R

predict as

R \to 0

(an ideal wire) with a fixed current

I

The voltage across it tends to zero:

V = I \times 0 = 0

. An ideal wire drops no voltage, which is why we treat connecting wires as having no resistance.

Edge case: What happens to

I

R \to \infty

(an open switch / broken wire) at fixed

V

The current tends to zero:

I = V / R \to 0

. Infinite resistance is a break in the circuit — the push exists but nothing can flow.

Edge case: If a battery is connected the "wrong way round", what happens to the point on the V–I graph?

Both

V

and

I

flip sign, so the operating point moves from the first quadrant to the third along the same line — the resistance is unchanged, only the direction of flow reverses.

Edge case: If you apply the same voltage to a resistor and to a filament lamp, why might their currents differ even at equal "rated" resistance?

The lamp's

R

climbs as it heats, so under load its effective resistance is higher than its cold value, giving less current than a truly ohmic resistor of the same cold

R

Edge case: A resistor obeys

V = I R

; at a fixed voltage, does a smaller

R

dissipate more or less power?

More. Power is

P = V I = V^{2} / R

(see Power in Circuits (P = VI)); at fixed

V

, shrinking

R

raises both current and power.

Recall One-line summary of every trap above

$R = V / I$ defines resistance; Ohm's Law is the extra fact that $R$ stays constant (and positive, in every quadrant) for ohmic conductors at fixed temperature. Confuse "definition" with "law", forget unit conversions, read a negative sign as an error, or swap across (V) and through (I), and you fall into a trap on this page.

Connections

Parent: Ohm's Law — the home note these traps drill.
Non-ohmic Components — where the straight-line assumption legitimately fails (orange curve).
Resistance — why geometry and temperature set $R$ .
Series and Parallel Circuits — which voltage to use for which component.
Power in Circuits (P = VI) — the edge case on energy dissipation.