13.4 Electrical Measuring Devices: Voltmeter, Ammeter, and Wattmeter

13.4 Electrical Measuring Devices: Voltmeter, Ammeter, and Wattmeter

NCEES lists "Measuring devices (e.g., voltmeter, ammeter, wattmeter)" as one of five Basic Electrical Engineering sub-topics. It is examined conceptually — connection topology, ideal internal impedance, and loading error — with occasional short calculations. Small topic, quick points.

The Two Connection Rules

Memorize the pairing: voltmeter — parallel — high resistance and ammeter — series — low resistance. Everything else in the topic follows from it.

The Consequences of Getting It Wrong

Loading Error

A real meter changes the circuit it measures. The magnitude of the disturbance depends on the ratio of meter resistance to circuit resistance.

Voltmeter Loading

A voltmeter of resistance $R_V$ across a resistor $R$ makes the effective resistance:

$R_{\text{eff}} = \frac{R\,R_V}{R + R_V}$

The reading is low, and the fractional error is approximately:

$\text{error} \approx -\frac{R_{\text{Th}}}{R_{\text{Th}} + R_V}$

where $R_{\text{Th}}$ is the Thevenin resistance seen by the meter.

Worked Example: Voltmeter Loading in a Divider

Two 1.0 MΩ resistors form a divider across 100 V. A voltmeter with $R_V = 10$ MΩ measures across the lower resistor. What does it read?

True value (no meter): $V = 100 \times \frac{1.0}{1.0+1.0} = 50.0$ V.

With the meter, the lower resistor is paralleled by 10 MΩ:

$R_{\text{eff}} = \frac{1.0(10)}{1.0+10} = 0.909\ \text{M}\Omega$

$V_{\text{read}} = 100 \times \frac{0.909}{1.0+0.909} = 100 \times \frac{0.909}{1.909} = \boxed{47.6\ \text{V}}$

A 4.8% error from a meter whose resistance is 10× the circuit resistance. This is why high-impedance circuits demand high-impedance meters: a modern digital multimeter at 10 MΩ is adequate here, while an old 20 kΩ/V analog meter would read about 2 V and be worthless.

The rule of thumb: keep $R_V \ge 100 R_{\text{Th}}$ for error under 1%. In low-impedance power circuits ($R_{\text{Th}}$ of a few ohms) loading is negligible; in sensor and instrumentation circuits it dominates.

Range Extension: Shunts and Multipliers

Ammeter Shunt (parallel)

To extend an ammeter of full-scale current $I_m$ and resistance $R_m$ to measure $I_{\text{total}}$:

$R_{\text{sh}} = \frac{I_mR_m}{I_{\text{total}} - I_m} = \frac{R_m}{n-1}, \qquad n = \frac{I_{\text{total}}}{I_m}$

Voltmeter Multiplier (series)

To extend a movement of full-scale voltage $V_m$ to measure $V_{\text{total}}$:

$R_{\text{mult}} = R_m\left(\frac{V_{\text{total}}}{V_m} - 1\right) = R_m(n-1)$

Note the symmetry: the shunt divides $R_m$ by $(n-1)$ and sits in parallel; the multiplier multiplies $R_m$ by $(n-1)$ and sits in series. Both follow directly from the connection rules.

For high AC currents and voltages, instrument transformers replace shunts and multipliers: a current transformer (CT) steps current down to a 5 A or 1 A secondary, and a potential transformer (PT) steps voltage down to 120 V. A CT secondary must never be open-circuited while primary current flows — the core saturates and dangerous voltages appear at the terminals.

The Wattmeter

A wattmeter has two coils and reads the true average power, automatically including the phase relationship:

$P = \frac{1}{T}\int_0^T v(t)i(t)\,dt = V_{\text{rms}}I_{\text{rms}}\cos\theta$

The wattmeter's defining advantage. On an AC circuit, a voltmeter reading times an ammeter reading gives apparent power $S = VI$ in volt-amperes — not real power. The wattmeter multiplies the instantaneous values before averaging, so it captures $\cos\theta$ and reports watts.

$\text{Power factor} = \frac{\text{wattmeter reading}}{\text{voltmeter} \times \text{ammeter}} = \frac{P}{S} = \cos\theta$

This is the standard exam item on this sub-topic, and it also gives you a practical way to determine power factor from three instruments.

Worked Example: Power Factor from Three Meters

A single-phase motor circuit reads 240 V, 18 A, and 3{,}240 W. Find the power factor, apparent power, and reactive power.

$S = VI = 240(18) = 4{,}320\ \text{VA}$

$pf = \cos\theta = \frac{P}{S} = \frac{3{,}240}{4{,}320} = \boxed{0.75\ \text{lagging}}$

$\theta = \arccos(0.75) = 41.4°$

$Q = S\sin\theta = 4{,}320\sin(41.4°) = 4{,}320(0.661) = 2{,}856\ \text{VAR}$

Check with the power triangle: $\sqrt{P^2+Q^2} = \sqrt{3{,}240^2 + 2{,}856^2} = \sqrt{1.050\times10^7 + 8.157\times10^6} = \sqrt{1.866\times10^7} = 4{,}320$ VA ✓

Trap: assuming $P = VI = 4{,}320$ W overstates real power by 33%, and a motor sized on that figure would be a third too large. Only the wattmeter reading is power.

Two-Wattmeter Method for Three-Phase Power

Any three-wire three-phase load — balanced or unbalanced, wye or delta — has its total real power measured by just two wattmeters:

$P_{\text{total}} = W_1 + W_2$

For a balanced load, the two readings also yield the power factor:

$\tan\theta = \sqrt{3}\,\frac{W_1 - W_2}{W_1 + W_2}$

The negative reading at low power factor surprises candidates but is physically correct, and the total is still $W_1 + W_2$ with the sign respected.

True-RMS vs. Average-Responding Meters

Average-responding meters are calibrated by multiplying the rectified average by the sine wave's form factor, 1.11. On the distorted, harmonic-rich currents drawn by variable-frequency drives, switch-mode supplies, and LED lighting, they can read 30–40% low. Where harmonics are present — which is most modern electrical work — a true-RMS instrument is required.