Voltmeter Loading in a Voltage Divider
A student assembles a voltage divider: two 10.0 kΩ resistors in series across a 12.0 V battery of negligible internal resistance. Theory says the midpoint must sit at 6.00 V. She measures it with an analog moving-coil voltmeter whose resistance on its 10 V range is 20 kΩ, and reads 4.8 V. Concluding that a resistor is faulty, she replaces both - the reading stays at 4.8 V. A demonstrator points out that a voltmeter is a resistor, not an observer, and that on the 10 V range this meter draws 50 μA at full scale (i.e. 2 kΩ per volt). The student must now decide whether the discrepancy is a circuit fault or a measurement artefact, how large the artefact is, and what meter specification would bring the error below 2 % of the true 6.00 V value. The battery and both resistors are known good, and the meter is calibrated.
1. Show that 4.8 V is exactly what the loaded divider should give, so nothing is faulty.
Read the worked answer
The voltmeter is a 20 kΩ resistance in parallel with the 10 kΩ lower arm:
R_p = (10 × 20)/(10 + 20) = 6.67 kΩ Total circuit = 10.0 + 6.67 = 16.67 kΩ; I = 12.0/16.67 k = 0.720 mA Reading = 0.720 mA × 6.67 kΩ = 4.80 V
This matches the observation exactly, so no component is faulty - the meter has loaded the divider. The error is 4.80 − 6.00 = −1.20 V (−20 %), and the reading is always low when a finite-resistance meter is placed across the lower arm. Replacing resistors cannot help; only the meter's resistance can.
2. What minimum voltmeter resistance keeps the error below 2 % on this divider?
Read the worked answer
Target reading ≥ 5.88 V (2 % below 6.00 V). Let x be the parallel combination of the meter R_m with the 10 kΩ arm:
12x/(10 + x) ≥ 5.88 → 12x ≥ 58.8 + 5.88x → x ≥ 9.61 kΩ 1/x = 1/10 + 1/R_m → 1/R_m ≥ 1/9.61 − 1/10 = 0.00406 → R_m ≥ 246 kΩ
So the meter needs roughly ≥ 250 kΩ on this range (≥ 25 kΩ/V); a 2 kΩ/V meter is far too coarse. Rule of thumb: for a loading error under about 1 %, the meter resistance should exceed the resistance across which it is connected by a factor of ~100.
3. She repeats the measurement with a digital meter of 10 MΩ input resistance. Predict the reading.
Read the worked answer
Now R_p = (10 kΩ × 10 MΩ)/(10 kΩ + 10 MΩ) ≈ 9.99 kΩ, barely different from 10.0 kΩ.
Total ≈ 19.99 kΩ; I = 12.0/19.99 k = 0.600 mA; reading = 0.600 × 9.99 = 5.997 V.
Error ≈ 0.05 %, negligible. This confirms the earlier discrepancy was purely a loading artefact: the same circuit gives 4.80 V, 5.997 V or 6.00 V depending only on how heavy the measuring instrument is relative to the circuit.