Voltmeter Loading a High-Resistance Divider
A student builds a voltage divider from two nominal 1.0 MΩ, ±5% carbon-film resistors in series across a regulated 12.00 V supply and expects 6.00 V across the lower resistor. A digital multimeter, specified as having 10 MΩ input resistance on every DC voltage range, reads 5.71 V. The student's partner insists this proves at least one resistor is out of tolerance and wants to replace both. As a quick test they substitute two 1.0 kΩ, ±5% resistors, and the same meter now reads 5.99 V. The supply is independently verified at 12.00 V, both resistor pairs are checked on a calibrated 4½-digit ohmmeter and agree with nominal to within 1%, and the meter agrees with a second meter on the 1.0 kΩ divider. The pair must decide whether the 1.0 MΩ network is faulty and how much confidence they should place in ideal-divider arithmetic in future measurements.
1. Show whether the 5.71 V reading is consistent with two good 1.0 MΩ resistors, and state how much the meter's own resistance matters here.
Read the worked answer
The meter has to be treated as a circuit element, not an observer. On its 20 V range it presents 10 MΩ in parallel with the lower 1.0 MΩ resistor:
R_p = (1.0 MΩ × 10 MΩ)/(1.0 MΩ + 10 MΩ) = 0.909 MΩ
The divider is now 1.0 MΩ in series with 0.909 MΩ across 12.00 V:
V = 12.00 × 0.909/(1.0 + 0.909) = 12.00 × 0.4762 = 5.71 V
That is exactly the observed value. Two perfect resistors predict the reading, so no fault is required and the partner's inference is unsupported.
The size of the effect is set by the ratio of the source resistance to the meter resistance. The divider's Thévenin resistance seen at the measurement node is 1.0 MΩ ∥ 1.0 MΩ = 0.50 MΩ, which is 5% of 10 MΩ - and the loading error is indeed about 5%, the gap between 6.00 V and 5.71 V. The usual rule is that a voltmeter's input resistance should be at least 100 times the Thévenin resistance of the point being probed to keep loading under about 1%. At 10:1 this circuit fails that test by a factor of five, which is why the error is a visible 5% rather than a rounding-level annoyance.
2. Using only the two readings and the known resistor values, determine the meter's input resistance as far as the data allow, and explain why the 1.0 kΩ trial contributes essentially nothing.
Read the worked answer
Let R_m be the meter resistance and R₂ = 1.0 MΩ. The node voltage is V = 12 R_eff/(R₁ + R_eff) with R_eff = R₂R_m/(R₂ + R_m).
Solving 5.71 = 12R_eff/(1.0 + R_eff) in megohms gives R_eff = 1.0 × 5.71/(12 − 5.71) = 0.909 MΩ.
Then 1/R_m = 1/0.909 − 1/1.0 = 1.100 − 1.000 = 0.100 MΩ⁻¹, so R_m = 10 MΩ - in exact agreement with the specification. The 'bad' measurement was in fact an accidental but quite precise measurement of the meter.
With the 1.0 kΩ pair, R_eff = 1.0 kΩ ∥ 10 MΩ ≈ 0.99990 kΩ, giving 5.9995 V. Loading shifts the reading by 0.005%, well below both the display resolution and the resistors' own 1% tolerance, so it carries no information about R_m. Sensitivity scales as R_source/R_m; to measure a 10 MΩ resistance you need a source resistance comparable to it. The 1.0 kΩ trial is a useful null check - it shows the meter is fine and the 5% supply is fine - but it cannot measure the meter.
3. Give two practical fixes that would let the student measure the 1.0 MΩ divider accurately, and note one cost of each.
Read the worked answer
Fix 1 - scale the resistors down. Two 1 kΩ resistors give the same 6.00 V division with loading near 0.01%. Cost: the divider then draws 12 V / 2 kΩ = 6 mA instead of 6 µA, dissipating 36 mW in each resistor (acceptable for 0.25 W parts) and loading the source. That is fine for a bench supply and unacceptable for a high-impedance sensor or a small battery, so the fix trades a measurement problem for a source-loading problem.
Fix 2 - raise the instrument's input impedance. A 10:1 FET probe, an electrometer input (10¹² Ω or better), or a unity-gain op-amp buffer at the node reduces loading to unmeasurable levels. Cost: extra equipment, and the buffer brings its own input offset voltage and bias current that must be characterised before the reading can be trusted at the millivolt level.
There is a third, purely computational fix: correct the reading using the known R_m, V_true = V_meas (R₁ + R_p)/R_p with R_p = R₂ ∥ R_m. It costs nothing but is fragile - it fails silently if the meter autoranges to a range with different input resistance, or if the meter is swapped - and it teaches nothing about the circuit the way the other two fixes do.