A potential divider that reads 4.80 V instead of 6.00 V
Two 10.0 kΩ resistors are connected in series across a 12.0 V bench supply whose internal resistance is negligible, forming a potential divider, so the p.d. across either resistor should be 6.00 V. A student measures the p.d. across the lower resistor with a handheld digital multimeter on its 20 V range and reads 4.80 V. Suspecting the supply, the student measures the supply terminals directly and reads 12.0 V. Both resistors are then removed and measured with the same meter on its 200 kΩ resistance range; each reads 9.98 kΩ. The student concludes that the divider is 'losing 1.2 V per resistor' and that the lower resistor must be faulty. The technician disagrees, says the circuit is behaving exactly as it should, and asks the student to work out what the meter is doing to the circuit before any component is replaced.
1. From the 4.80 V reading, calculate the input resistance of the meter and explain why this makes the meter the problem rather than the resistor.
Read the worked answer
A real voltmeter has a finite resistance Rm and sits in parallel with the resistor it spans. The upper 10 kΩ is unchanged; the lower branch becomes 10 kΩ ∥ Rm. The measured ratio is
4.80 / 12.0 = 0.400 = R_par / (10.0 kΩ + R_par),
so 0.600 R_par = 4.00 kΩ and R_par = 6.67 kΩ.
Then 1/R_par = 1/10.0 kΩ + 1/Rm, giving 1/Rm = 1.50×10⁻⁴ − 1.00×10⁻⁴ = 0.50×10⁻⁴ Ω⁻¹, so Rm = 20 kΩ.
That is a plausible value for an inexpensive meter on a low range (and for an analogue meter of modest Ω/V figure), but it is far too low to be an acceptable load on a 5 kΩ source. The reading is exactly what a 20 kΩ meter should give on this circuit; no component is faulty. The '1.2 V per resistor' loss is the loading error appearing on the one branch being probed. Note also that checking the resistors in resistance mode proves only that they are 9.98 kΩ - it says nothing about the voltmeter's input resistance.
2. Predict the reading if the lower resistor were replaced by 100 kΩ, keeping the same meter, and explain what is counter-intuitive about the result.
Read the worked answer
New parallel combination: 100 kΩ ∥ 20 kΩ = (100 × 20)/120 = 16.67 kΩ.
V = 12.0 × 16.67 / (10.0 + 16.67) = 12.0 × 0.625 = 7.50 V.
The true value is still 6.00 V, so the error has grown from −1.20 V (−20%) to +1.50 V (+25%). The counter-intuitive part is that the reading moves further from the truth even though it has moved closer to the supply voltage and 'looks more sensible'. The reason is the ratio Rm / R, where R is the resistance being spanned: it was 2.0 (bad), and is now 0.2 (worse). Loading error is governed by the meter resistance relative to the resistance across which you measure, not by the absolute number on the display. This is the practical argument for the rule of thumb that a voltmeter should have an input resistance at least 100 times the Thévenin resistance at the measurement point (here 10 kΩ ∥ 10 kΩ = 5 kΩ, so ≥ 500 kΩ).
3. Recommend a measurement strategy that gives the true 6.00 V and justify the criterion you use.
Read the worked answer
Use a meter whose input resistance is at least 100 × R_Th at the measurement node. Here R_Th = 10 kΩ ∥ 10 kΩ = 5.0 kΩ, so the meter must exceed about 500 kΩ; a 10 MΩ DMM loads the node by 5 kΩ/10 MΩ ≈ 0.05%, i.e. 3 mV - negligible.
Alternatives if only the 20 kΩ meter is available: (i) calculate the correction using the known Rm - with the measured value and Rm you can invert the divider equation, though the correction amplifies any reading error and is only valid if Rm is accurately known and stable across ranges; (ii) buffer the node with a unity-gain op-amp follower, whose input bias current and offset dominate only at much higher source impedances; (iii) use a null method (potentiometer or bridge), where no current is drawn at balance.
What must not be concluded is that a component is faulty. The diagnostic ordering matters: check the instrument's loading first, because it is the cheapest hypothesis to test and it explains both numbers without invoking a fault.