Read multimeter measurements (V, I, R): Worked examples

Everything below rests on the parent note the parent topic and on one law you already know: Ohm's Law, written $V = I \times R$ ("voltage equals current times resistance"). Every meter mode is just this law solved for a different letter.

The scenario matrix

Before any example, here is the full grid of situations this topic can throw at you. Each row is a case class; the last column names the example that covers it.

Cell	Case class	What makes it tricky	Covered by
A	Normal voltage read, positive	Read digits × prefix	Example 1
B	Sign flip — probes reversed (voltage)	Screen shows minus, magnitude same	Example 2
C	Manual range / prefix trap	"0.47" is NOT 0.47 Ω	Example 3
D	Current in series, cross-check	Predict with Ohm's law, verify	Example 4
E	Degenerate: open circuit	Ammeter reads 0, ohmmeter reads OL/∞	Example 5
F	Degenerate: dead short	$R \to 0$ , $V \to 0$ , current spikes	Example 6
G	Live-circuit resistance error	Parallel paths lower the reading	Example 7
H	Limiting: loading effect	High source resistance corrupts V	Example 8
I	Real-world word problem	Choose mode + jack yourself	Example 9
J	Exam twist — combine everything	Series drop, prefix, sign, sanity	Example 10
K	Sign flip — probes reversed (current)	Negative current, same magnitude	Example 11
L	Mode: AC vs DC	Wrong mode gives nonsense	Example 12

We now walk each cell. Read the Forecast and try to guess before scrolling to the steps.

Cell A — Normal positive voltage

Worked example Example 1 — A single resistor across a 9 V supply

A $9 V$ battery feeds one resistor. You set the dial to DCV, red probe on the top of the resistor, black on the bottom. Screen: 9.0.

Forecast: What voltage should the resistor show if it is the only component?

Identify the mode. DCV = direct-current voltage — steady, not oscillating. Why this step? A battery is DC; ACV would read near zero on a steady source.
Recognise the circuit. One resistor, one source, one loop. All the battery's "push" lands across that single resistor. Why this step? In a single-element loop the source voltage equals the element voltage — nothing else to share it with.
Read digits × prefix. Screen shows $9.0$ with no prefix letter, so it is $9.0 V$ . Why this step? The number is meaningless until you attach the unit implied by the mode and prefix; here there is no prefix, so volts stand as-is.

Verify: The resistor is the only place voltage can drop, so it must equal the source: $V_{R} = 9 V$ . ✔ Units: volts across two points — correct for a difference.

Cell B — Sign flip (probes reversed, voltage)

Worked example Example 2 — The screen shows a minus sign

Same $9 V$ battery. This time you accidentally put black on top and red on bottom. Screen: −9.0.

Forecast: Is your battery broken? Is the reading wrong?

Interpret the sign. The DMM defines the reading as (voltage at red probe) − (voltage at black probe). Why this step? Voltage is a difference; swapping which point is "reference" flips the sign, exactly like measuring a hill's height as +5 m going up or −5 m going down.
Compare magnitudes. $∣ - 9.0 ∣ = 9.0 V$ , identical to Example 1. Why this step? The physical push is unchanged; only the bookkeeping direction changed.
Decide. Nothing is broken. The minus tells you red is at the lower potential. Why this step? Diagnosing a "fault" that is really a wiring convention prevents you from throwing away a healthy battery.

Verify: Flip the probes back → +9.0 V returns. Magnitude preserved, only sign toggled. ✔

Cell C — The prefix / range trap

Worked example Example 3 — "0.47" on the 20 kΩ range

A resistor is out of the circuit. Manual-range dial is on 20 k. Screen: 0.470.

Forecast: Is this a 0.47 Ω resistor?

Read the range, not just the digits. The dial says the full-scale value is $20 k Ω$ , so the unit of the whole reading is kilo-ohms. Why this step? On a manual meter the dial position IS the prefix (see Metric Prefixes and Engineering Notation).
Attach the unit. $0.470 \times 1 k Ω = 0.470 k Ω$ . Why this step? The kilo-ohm unit from the dial must be multiplied onto the raw digits, or the reading is off by a factor of 1000.
Convert to plain ohms. $0.470 k Ω = 0.470 \times 1 0^{3} Ω = 470 Ω$ . Why this step? Colour codes and datasheets quote plain ohms, so we convert to compare against them.

Verify: Colour bands yellow-violet-brown = $47 \times 1 0^{1} = 470 Ω$ (see Resistor Colour Codes). ✔ Matches — it is 470 Ω, not 0.47 Ω (off by a factor of 1000).

Cell D — Current in series, cross-checked

Worked example Example 4 — Current through an LED branch

A $270 Ω$ resistor and an LED (forward drop $1.6 V$ ) run from $5 V$ . You measure current.

Forecast: How many mA should flow?

Find the resistor's voltage. The LED eats $1.6 V$ , leaving the resistor $5 - 1.6 = 3.4 V$ . Why this step? In series the source voltage splits among elements; whatever the LED doesn't take, the resistor does.
Apply Ohm's law for current. $I = \frac{V _{R}}{R} = \frac{3.4}{270} = 0.01259 A$ . Why this step? We know $V_{R}$ and $R$ , so $I$ falls straight out of $V = I R$ .
Read the prefix. $0.01259 A = 12.6 mA$ .
Measure to confirm. Move red to the mA jack, break the wire, insert the meter in series. Screen: 12.7 mA. Why this step? $12.6 mA$ sits comfortably inside the mA jack's range (typically good to a few hundred mA), so the mA jack is the correct, fused choice here — no need for the high-current 10 A jack.

Verify: Predicted $12.6 mA$ vs measured $12.7 mA$ — agreement within $1%$ . ✔ The tiny gap is rounding in the LED drop; the meter is trustworthy.

Cell E — Degenerate: open circuit

Worked example Example 5 — A broken wire

A wire in the loop is snapped. You test it two ways.

Forecast: What does the ammeter read? What does the ohmmeter read?

Ammeter across the break (in series). No complete path exists, so no charge can flow: $I = 0 mA$ . Why this step? Current needs a closed loop; an open loop carries none.
Ohmmeter across the break. The meter tries to push its tiny test current $I_{t es t}$ , but the gap blocks it. Effective resistance $\to \infty$ . Why this step? $R = V_{m e a s u r e d} / I_{t es t}$ ; with $I_{t es t}$ forced to $\approx 0$ through an open gap, $R$ blows up.
Read the screen. It shows an overload indication — not a number. Why this step? An infinite resistance cannot be printed as a digit, so meters flag it with a special symbol; you must recognise the flag rather than mistake it for a value.

Verify: As $R \to \infty$ , $I = V / R \to 9/\infty = 0$ . The two views agree: open circuit ⇒ zero current, infinite resistance. ✔

Cell F — Degenerate: dead short

Worked example Example 6 — A wire straight across the resistor

Someone bridges the $1 k Ω$ resistor with a bare wire ( $\approx 0 Ω$ ).

Forecast: What happens to the resistance, the voltage across it, and the current?

Ohmmeter across the short. The wire's resistance is essentially $0 Ω$ ; the screen reads $\approx 0.0 Ω$ (often with a continuity beep). Why this step? A short is the opposite extreme of an open — it invites current instead of blocking it.
Voltage across the short. $V = I \times R = I \times 0 = 0 V$ . Why this step? With zero resistance there is nothing for voltage to "develop across."
Current through the loop. With the resistor bypassed, the only thing left to limit current is the battery's own internal resistance $R_{in t}$ . A typical AA-cell internal resistance is a fraction of an ohm — take $R_{in t} = 0.3 Ω$ as a representative value. Then $I = 9/0.3 = 30 A$ — a dangerous spike. Why this step? This shows why a short is hazardous: once the external resistor is gone, the tiny (but non-zero) internal resistance is all that stops the current running away.

Verify: Limiting behaviour: as $R \to 0$ , $V_{R} = I R \to 0$ and $I = V / R_{in t}$ blows up. ✔ Both consistent with a dead short.

Cell G — Live / in-circuit resistance error

Worked example Example 7 — The "broken" 1 kΩ resistor

A resistor marked $1 k Ω$ reads only 220 Ω on the Ω dial — still soldered in, powered off. A second $270 Ω$ resistor happens to sit in parallel with it.

Forecast: Is the resistor faulty, or is the reading being pulled down?

Spot the parallel path. The ohmmeter's test current can flow through both resistors, so it measures the parallel combination, not just one. Why this step? Parallel resistors always combine to less than the smallest one.
Compute the parallel value. $R_{∥} = \frac{1000 \times 270}{1000 + 270} = \frac{270000}{1270} = 212.6 Ω$ . Why this step? If this roughly matches the $220 Ω$ reading, the "fault" is really the parallel path.
Compare. $212.6 Ω \approx 220 Ω$ (meter tolerance). The resistor is fine. Why this step? Matching the computed parallel value to the reading is what proves the low number is expected physics, not a dead component — saving you from replacing a good part.

Verify: Lift one leg (isolate) and re-measure → $\approx 1 k Ω$ . ✔ Rule: always isolate before reading Ω.

Cell H — Limiting case: the loading effect

Worked example Example 8 — When even a 10 MΩ meter lies

You measure the midpoint of a divider made of two $10 M Ω$ resistors across $10 V$ . True midpoint voltage should be $5 V$ . Your meter's input resistance is $R_{in} = 10 M Ω$ .

Forecast: Will the meter read exactly 5 V?

Model the meter as a resistor. In voltage mode the DMM is a big resistor $R_{in} = 10 M Ω$ placed across the lower resistor. Why this step? A real voltmeter is not infinite; it steals a little current (see Loading Effect and Meter Accuracy).
Combine meter with lower resistor (parallel). $R_{l o w ∥ m e t er} = \frac{10 \times 10}{10 + 10} = 5 M Ω$ . Why this step? The meter and the lower $10 M Ω$ now share the current, dropping the effective lower resistance.
Recompute the divider. $V_{r e a d} = 10 \times \frac{5}{10 + 5} = 10 \times \frac{5}{15} = 3.33 V$ . Why this step? The divider now favours the (unchanged) upper $10 M Ω$ , so the midpoint sags.

Verify: True value $5 V$ , meter reads $3.33 V$ — a $33%$ error. ✔ Limiting insight: loading only matters when source resistance is comparable to $R_{in}$ . For a low-resistance source it vanishes.

The figure below makes the sag visible. Its vertical axis is the measured midpoint voltage in volts (0 to 6 V, with gridline ticks). The left green bar is the true, unloaded midpoint at $5 V$ ; the right red bar is what the meter actually reports, $3.33 V$ ; the yellow dashed line marks the ideal $5 V$ level; and the blue double-arrow measures the $1.67 V$ drop the meter itself caused by loading. Each bar is also labelled with its numeric value printed on top, so the reading is legible without relying on colour alone.

Figure — Read multimeter measurements (V, I, R)

Cell I — Real-world word problem

Worked example Example 9 — "The phone charger feels weak"

A USB charger should output $5 V$ and deliver up to $2 A$ . Under a $2.5 Ω$ load it seems dim. You have one multimeter. Diagnose it.

Forecast: Which mode, which jack, and what number confirms a healthy charger?

Choose voltage first (safest, non-invasive). Dial to DCV, probes across the output. This never breaks the circuit. Why this step? Voltage is measured in parallel; you risk nothing.
Predict the healthy current. If output holds $5 V$ across $2.5 Ω$ : $I = 5/2.5 = 2 A$ . Why this step? This tells you what the ammeter should show if the charger is fine.
Measure current in series. Red to the 10 A jack (2 A exceeds the mA fuse), break the load line, insert the meter. Screen: 2.0 A. Why this step? $2 A$ is far above the mA range; using the mA jack would blow its fuse.

Verify: $V = 5 V$ steady and $I = 2 A$ ⇒ delivering $P = V \times I = 5 \times 2 = 10 W$ , the charger's rated power. ✔ The "weak" feeling is the load, not the charger.

Cell J — Exam twist (everything at once)

Worked example Example 10 — Series drop + prefix + sign

Two resistors in series across $12 V$ : $R_{1} = 2 k Ω$ , $R_{2} = 4 k Ω$ . You measure across $R_{2}$ but connect black on top, red on bottom. Meter is on the 20 V range.

Forecast: What exact number (sign and all) appears on screen?

Find the loop current. Series resistances add: $R_{t o t} = 2 k + 4 k = 6 k Ω$ . Then $I = 12/6000 = 2 mA$ . Why this step? One current flows through a series chain; find it once.
Voltage across $R_{2}$ . $V_{2} = I \times R_{2} = 0.002 \times 4000 = 8 V$ . Why this step? Ohm's law on the single element gives its share of the $12 V$ .
Apply the sign. Probes reversed ⇒ the screen shows $- 8 V$ . Why this step? Reading = red − black; with red at the lower point the difference is negative.
Check the range. $20 V$ range comfortably fits $8 V$ ; digits read directly in volts.

Verify: $V_{1} + V_{2} = I R_{1} + I R_{2} = 0.002 (2000) + 0.002 (4000) = 4 + 8 = 12 V$ = source. ✔ Sign is $-$ , magnitude $8 V$ .

Cell K — Sign flip during current measurement

Worked example Example 11 — A negative current reading

Back to the LED branch of Example 4 (predicted $12.6 mA$ ). You insert the ammeter in series but with the probes swapped: current enters the black lead and exits the red lead. Screen: −12.7 mA.

Forecast: Is the current really flowing "backwards," or is something else going on?

Recall the meter's sign rule. A DMM in current mode calls flow positive when it enters the red lead and exits the black lead. Why this step? Just like voltage, current sign is a convention about which lead is the reference direction — not a claim about physics.
Interpret the minus. Current is actually entering black and exiting red, i.e. opposite to the meter's assumed positive direction, so it prints a minus. Why this step? Understanding that the minus is a direction label stops you from wrongly concluding the LED is wired backwards.
Compare magnitudes. $∣ - 12.7 ∣ = 12.7 mA$ — identical to Example 4. Why this step? The physical amount of charge per second is unchanged; only the bookkeeping direction flipped.

Verify: Swap the leads back → $+ 12.7 mA$ returns, matching the Ohm's-law forecast $12.6 mA$ . ✔ Magnitude preserved, only the sign toggled — exactly the current-mode twin of the voltage sign flip in Example 2.

Cell L — Mode: AC vs DC

Worked example Example 12 — The same probes, two different modes

A wall adapter's transformer output is $12 V$ AC (root-mean-square). You leave the meter on DCV by mistake. Screen: 0.1. Then you switch to ACV. Screen: 12.0.

Forecast: Why did DCV read almost nothing while ACV read 12?

Recall what AC does. AC (alternating current) voltage swings positive then negative many times a second; its average over a full cycle is $\approx 0$ . Why this step? A source's waveform decides which mode reads it correctly.
DCV averages. The DCV mode reports the average level. Averaging a symmetric up-and-down swing gives $\approx 0$ , so the screen shows a tiny leftover like $0.1 V$ . Why this step? This explains the near-zero nonsense reading — the mode threw away the very quantity you wanted.
ACV measures the effective (RMS) size. The ACV mode is built to report the root-mean-square value, the "effective" size of the swing, giving the true $12.0 V$ . Why this step? Only the matching mode reports a meaningful number; the digits are correct on the $20 V$ AC range with no extra prefix.

Verify: DC average of a symmetric AC wave $\approx 0 V$ ✔; ACV reports RMS $= 12 V$ ✔. Rule: wall/adapter AC → ACV; battery/regulated supply → DCV. Wrong mode gives near-zero or nonsense.

Recall Feynman: the whole matrix in one breath

Voltage is a difference, so it can go plus or minus depending on which whisker you call "home" — and current has the very same plus/minus quirk depending on which lead the flow enters. Current needs a full loop — snap it open and you get zero, short it out and you get a scary flood. Resistance only tells the truth when the part is alone and unpowered, because sneaky parallel neighbours always drag the number down. Even a "perfect" 10-million-ohm voltmeter lies a little when the thing it measures is itself made of millions of ohms. And you must match the mode to the waveform — a wobbling AC signal read on DC just averages away to nothing. Read the letter next to the number, watch the sign, pick AC or DC, and always ask "what path is the electricity actually taking?"

Active-recall

A meter reads "0.47" on the 20 kΩ range — what is the resistance?

0.47 \times 1 k Ω = 470 Ω

, not 0.47 Ω.

Screen shows −9.0 V. Is the battery broken?

No — the probes are reversed; magnitude is 9 V, the minus is a direction label.

The ammeter reads −12.7 mA. What does the minus mean?

Current is entering the black lead instead of the red; magnitude is 12.7 mA, direction is just reversed.

Ohmmeter across an open circuit shows OL / a lone 1 — what is that?

An over-limit / infinite-resistance flag (conventions vary), meaning the resistance is too high to measure — NOT 1 Ω.

Why did the in-circuit resistor read 220 Ω instead of 1 kΩ?

A parallel path lowered it; isolate one leg to read the true value.

Why does a 10 MΩ voltmeter still misread a two-10-MΩ divider?

The meter's resistance is comparable to the source, so it loads the divider — reads 3.33 V instead of 5 V.

A 12 V AC adapter read on DCV shows ~0.1 V — why?

DCV reports the average, and a symmetric AC swing averages to ~0; switch to ACV to read 12 V (RMS).