Read multimeter measurements (V, I, R): Exercises — Hardware deep dive

Level 1 — Recognition

These test knowing which mode, which jack, which prefix — no arithmetic beyond scaling.

Recall Solution L1.1

A regulated DC adapter puts out a steady voltage that does not flip direction, so use DCV (the DCV arc on the dial in Figure s01).

DCV measures the average (steady) level → reads $\approx 12 V$ .
ACV looks for a back-and-forth swing. A steady DC line has (almost) no swing, so ACV reads $\approx 0$ or a small nonsense number. Answer: DCV; the wrong setting (ACV) reads near zero.

Recall Solution L1.2

On a manual meter the dial position is the prefix/unit. The $20 k$ range means the reading is in kilo-ohms. $R = 1.47 k Ω = 1.47 \times 1 0^{3} Ω === 1470 Ω ==$ Answer: $1470 Ω$ (about $1.5 k Ω$ ).

Recall Solution L1.3

An LED draws small current (a few $mA$ ), so use the VΩmA jack (the fused mA path).

COM is the shared return ("common") for every measurement, so the black probe stays there always.
The 10A jack is a separate, often unfused path for big currents — overkill here and riskier. Answer: red → VΩmA, black stays in COM.

Recall Solution L1.4

The battery is perfectly fine. A DMM reports the voltage of the red probe relative to the black (COM) probe. A minus sign just means the red probe is touching the lower-potential point — here you put red on the (−) terminal and black on the (+).

Fix / flip: swap the two probes (red → (+), black → (−)) and the screen reads +1.48.
Useful feature, not an error: the sign tells you the polarity of the two points, which is handy when you don't know which terminal is which. Answer: not faulty; the reading is $1.48 V$ with reversed probes — swap probes to get $+ 1.48$ .

Level 2 — Application

Now plug numbers into Ohm's Law $V = I R$ , solving for whichever letter the meter can't reach directly.

Recall Solution L2.1

WHAT: we know $V$ and $R$ , want $I$ . WHY this rearrangement: Ohm's law $V = I R$ solved for the unknown $I$ gives $I = V / R$ . $I = \frac{V}{R} = \frac{1.98 V}{330 Ω} = 0.006 A === 6 mA ==$ To convert: $0.006 A = 0.006 \times 1 0^{3} mA = 6 mA$ . Answer: $6 mA$ .

Recall Solution L2.2

WHY this formula: resistance mode uses $R = V_{m e a s u r e d} / I_{t es t}$ — it forces a known current and measures the voltage, so $R$ falls out of $V = I R$ . First put $I_{t es t}$ in amps: $100 μ A = 100 \times 1 0^{- 6} A = 1 \times 1 0^{- 4} A$ . $R = \frac{V _{m e a s u r e d}}{I _{t es t}} = \frac{0.45 V}{1 \times 1 0 ^{- 4} A} = 4500 Ω === 4.5 k Ω ==$ Answer: $4.5 k Ω$ .

Recall Solution L2.3

WHY: an ammeter never measures amps directly — it measures the voltage across a tiny known resistor and applies $I = V_{s h u n t} / R_{s h u n t}$ . Convert: $25 mV = 25 \times 1 0^{- 3} V = 0.025 V$ . $I = \frac{V _{s h u n t}}{R _{s h u n t}} = \frac{0.025 V}{0.10 Ω} === 0.25 A === 250 mA$ Answer: $0.25 A$ (i.e. $250 mA$ ).

Recall Solution L2.4

Compare the current to each range's ceiling.

$200 mA = 0.2 A$ ceiling — but you expect $\approx 2 A$ , which is $10 \times$ over. The current would rush through the mA path, exceed the $0.5 A$ fuse rating, and blow the fuse (screen may briefly read OL then go dead).
$10 A$ ceiling — comfortably above $2 A$ , so the reading fits. Answer: use the 10A jack on the 10A range. Rule: always choose a current range whose ceiling is above your expected current; when unsure, start on the highest range and step down.

Level 3 — Analysis

Here the reading disagrees with expectation and you must reason about why.

Recall Solution L3.1

Step 0 — before you even touch the dial: for any resistance reading you must power OFF the circuit and, ideally, lift one leg of the part so no live current flows into the meter. Here the part is not isolated (both legs still connected), which is exactly why the reading is wrong — that is the whole point of the exercise.

WHAT the meter's own test current sees — read Figure s02. In Figure s02 the meter's probes (plum arrows on the left) push the tiny test current $I_{t es t}$ into the top horizontal rail. From that rail the current can drop down two vertical bars to the bottom rail: the orange bar is $R_{1}$ (the $1 k Ω$ part you think you're measuring) and the teal bar is $R_{2}$ (the hidden neighbour). Because both bars connect the same top rail to the same bottom rail, they are in parallel — that is literally what "two paths joined at both ends" means. Watch the two small split-arrows $I_{1}$ and $I_{2}$ : the test current divides between the bars.

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

WHY $\frac{1}{R _{∥}} = \frac{1}{R _{1}} + \frac{1}{R _{2}}$ (built, not quoted). Both paths bridge the same two nodes, so they feel the same voltage $V$ . The current in each obeys Ohm's law: $I_{1} = V / R_{1}$ and $I_{2} = V / R_{2}$ . The test current splits and then re-merges, so the total current is $I_{t es t} = I_{1} + I_{2}$ . Divide that whole equation by $V$ : $\frac{I _{t es t}}{V} = \frac{I _{1}}{V} + \frac{I _{2}}{V} = \frac{1}{R _{1}} + \frac{1}{R _{2}} .$ But $V / I_{t es t}$ is exactly the resistance the meter reports, call it $R_{∥}$ , so $I_{t es t} / V = 1/ R_{∥}$ . That gives the rule — it is just "currents add" plus Ohm's law, nothing memorised.

Now solve for $R_{2}$ with $R_{∥} = 0.40 k Ω = 400 Ω$ and $R_{1} = 1000 Ω$ : $\frac{1}{R _{2}} = \frac{1}{R _{∥}} - \frac{1}{R _{1}} = \frac{1}{400} - \frac{1}{1000} = \frac{5 - 2}{2000} = \frac{3}{2000}$ $R_{2} = \frac{2000}{3} \approx== 667 Ω ==$ Answer: $R_{2} \approx 667 Ω$ . Lesson: the "broken" $1 k Ω$ is fine — the parallel neighbour stole some of the test current and dragged the reading below the true value. Always power OFF and isolate before trusting an Ω reading.

Recall Solution L3.2

Each branch sees the full $12 V$ (parallel branches share the source voltage).

Branch A current: $I_{A} = \frac{12}{100} === 0.12 A === 120 mA$ — this is what the meter shows.
Branch B current: $I_{B} = \frac{12}{300} = 0.04 A = 40 mA$ .
True total: $I_{A} + I_{B} = 0.12 + 0.04 === 0.16 A === 160 mA$ . Answer: meter reads $120 mA$ , but total is $160 mA$ . Lesson: an ammeter only counts the current in the branch it cuts into.

Recall Solution L3.3

On $2 k$ : reading is in kilo-ohms → $0.470 k Ω === 470 Ω ==$ . ✔ correct value.
On $200Ω$ : the range's maximum is $200 Ω$ . A $470 Ω$ resistor is bigger than the range can display, so the meter shows OL ("over-limit"). Answer: $470 Ω$ ; the $200Ω$ range overflows because $470 > 200$ . Lesson: pick a range whose top value is above what you're measuring.

Recall Solution L3.4

Not dangerous, unlike over-ranging current. In voltage mode the meter's input is a huge $\approx 10 M Ω$ resistor, so even $9 V$ pushes only $I = 9/ (10 \times 1 0^{6}) = 0.9 μ A$ — a harmless trickle. The OL simply means the number ( $9 V$ ) exceeds the selected ceiling ( $2 V$ ); nothing burns.

Fix: turn the dial up to the $20 V$ range (ceiling $20 > 9$ ) and read $\approx 9.0 V$ . Contrast with current: an over-range in current mode blows the fuse (L2.4) because that path is near-zero-resistance; an over-range in voltage mode is safe because that path is near-infinite-resistance. Answer: safe; move to the $20 V$ DCV range.

Level 4 — Synthesis

Combine modes and rules into one full measurement story.

Recall Solution L4.1

(a) Voltage across resistor — the supply splits between resistor and LED (series voltages add, from Series and Parallel Circuits): $V_{R} = V_{s u ppl y} - V_{L E D} = 5 - 1.8 === 3.2 V ==$ (b) Current — one series loop, so the resistor's current is the branch current: $I = \frac{V _{R}}{R} = \frac{3.2}{220} \approx 0.01455 A \approx== 14.5 mA ==$ (c) Power in the resistor — $P = V_{R} \cdot I = 3.2 \times 0.01455 \approx== 0.0466 W ==\approx 47 mW$ .

What each of the three meter modes shows. The multimeter is really three instruments (voltmeter / ammeter / ohmmeter). There is no separate "power" mode — power is never measured directly, it is computed from a V and an I reading, which is exactly what part (c) does. Here is each mode on this live-or-dead circuit:

DCV (voltmeter, power ON), across the resistor → $\approx 3.2 V$ . Across the LED it would read $\approx 1.8 V$ , and the two must sum to the $5 V$ supply ✔.
DC current mode (ammeter, power ON), in series — break the loop and insert the meter → $\approx 14.5 mA$ (choose a range whose ceiling exceeds this, e.g. the $200 mA$ range).
Ω (ohmmeter, power OFF, resistor isolated) — you must turn the supply off and lift one leg of the resistor, then the meter reads $\approx 220 Ω$ . Left in-circuit and live it would be wrong (and could blow the fuse). Answer: DCV $\to 3.2 V$ , current $\to 14.5 mA$ , Ω $\to 220 Ω$ ; multiply the first two to get the $\approx 47 mW$ of part (c).

Recall Solution L4.2

Convert current: $3.0 mA = 3.0 \times 1 0^{- 3} A$ . $R_{X} = \frac{V}{I} = \frac{6.0}{3.0 \times 1 0 ^{- 3}} = 2000 Ω === 2 k Ω ==$ Answer: $R_{X} = 2 k Ω$ . Both meter readings agree with Ohm's law, so X behaves as a plain resistor.

Level 5 — Mastery

Now the meter itself changes the answer — the loading effect. See Loading Effect and Meter Accuracy and Internal Resistance of Sources.

Recall Solution L5.1

WHY $R_{in}$ ends up in parallel with $R_{2}$ — read Figure s03. In Figure s03 the divider runs top-to-bottom: the orange bar is the upper resistor $R_{1}$ , the teal bar is the lower resistor $R_{2}$ , and the dot between them is the midpoint node whose true voltage is $5 V$ . A voltmeter measures across two points: here the midpoint node and the bottom (ground) node. But a voltmeter is not a magic window — internally it is a big resistor $R_{in}$ (the plum bar on the right), and to read the voltage it connects that resistor between exactly those two points (dashed plum lines). Now stare at what shares those two nodes: the lower divider resistor $R_{2}$ already spans midpoint→ground, and the voltmeter's $R_{in}$ now also spans midpoint→ground. Two resistors bridging the same pair of nodes = parallel (same reasoning we built in L3.1). So the moment you touch the probes, the lower arm of the divider silently becomes $R_{2} ∥ R_{in}$ .

Combine them with the parallel rule (product-over-sum form): $R_{2}^{'} = \frac{R _{2} \cdot R _{in}}{R _{2} + R _{in}} = \frac{1 \times 10}{1 + 10} M Ω = \frac{10}{11} M Ω \approx 0.909 M Ω$ Now it's a divider of $R_{1} = 1 M Ω$ (top) and $R_{2}^{'} = 0.909 M Ω$ (bottom). The midpoint voltage is the supply times the fraction of total resistance sitting in the lower arm: $V_{r e a d} = 10 \times \frac{R _{2}^{'}}{R _{1} + R _{2}^{'}} = 10 \times \frac{0.909}{1 + 0.909} = 10 \times \frac{0.909}{1.909} \approx== 4.76 V ==$ Answer: $\approx 4.76 V$ , not $5 V$ — a real loading error of about $0.24 V$ ( $\approx 4.8%$ ), because the source resistances are comparable to $R_{in}$ .

Recall Solution L5.2

Same set-up: $R_{in}$ sits in parallel with the $1 k Ω$ lower arm. $R_{2}^{'} = \frac{1 k \times 10 M}{1 k + 10 M} Ω = \frac{1 0 ^{10}}{1 0 ^{7} + 1 0 ^{3}} Ω \approx 999.9 Ω$ $V_{r e a d} = 10 \times \frac{999.9}{1000 + 999.9} \approx== 4.99975 V ==$ Answer: $\approx 5.00 V$ . Lesson: loading only matters when the circuit's own resistance is a sizeable fraction of $R_{in}$ . With $k Ω$ resistors and a $10 M Ω$ meter, the error is under $0.01%$ .

Recall Solution L5.3

WHY: the internal resistance $r$ is in series with the ideal cell (see Internal Resistance of Sources). Under load, the current is $I = \frac{V _{L}}{R _{l o a d}} = \frac{1.40}{10} = 0.14 A$ The "missing" voltage $V_{oc} - V_{L}$ is dropped across $r$ : $r = \frac{V _{oc} - V _{L}}{I} = \frac{1.50 - 1.40}{0.14} = \frac{0.10}{0.14} \approx== 0.714 Ω ==$ Answer: $r \approx 0.71 Ω$ . This is exactly the "small gap" the parent note's Example 4 waved at — now quantified.

Recall One-line summary of the whole ladder

Pick the right mode/jack/prefix (L1, and mind the polarity sign) → apply $V = I R$ for the missing letter, on a range whose ceiling beats your value (L2) → explain disagreements via parallel/branch reasoning (L3) → chain modes across a full circuit (L4) → account for the meter's own $R_{in}$ and the source's $r$ (L5).