The Voltage Divider That Would Not Split in Half
A student builds a voltage divider from two 10 MΩ, 5% carbon resistors in series across a fresh 9.00 V battery, expecting each resistor to take 4.50 V. The available digital multimeter is specified as having a 10 MΩ input resistance on its DC voltage ranges. When the student measures across the lower resistor, the display reads 3.00 V; measuring across the upper resistor gives 6.00 V. The two readings add to 9.00 V, and the battery itself measures 9.00 V, so the student concludes that one resistor must be faulty and asks for a replacement. The instructor asks the student to compute what the meter itself is doing to the circuit before changing any hardware.
1. Explain the readings quantitatively and determine whether either resistor is faulty.
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The meter is not an ideal voltmeter; it is a 10 MΩ resistor connected in parallel with whatever it measures.
Measuring across the lower 10 MΩ resistor puts 10 MΩ in parallel with it: R_p = (10 × 10)/(10 + 10) = 5.0 MΩ. The divider now consists of 10 MΩ (upper) in series with 5.0 MΩ (lower) across 9.00 V. Total resistance 15 MΩ, current 9.00/15 MΩ = 0.600 µA, and the voltage across the lower section is 0.600 µA × 5.0 MΩ = 3.00 V. That matches the display exactly.
Across the upper resistor, the meter loads the 10 MΩ upper resistor to 5.0 MΩ, leaving 10 MΩ below; by symmetry the reading is 6.00 V. The two readings sum to the source voltage because the meter's resistance is the same in both measurements.
No resistor is faulty. The 3.00 V and 6.00 V readings are exactly what an ideal 9.00 V source with two equal 10 MΩ resistors and a 10 MΩ meter must produce. The 5% resistor tolerance is irrelevant here; the error is 33%, far outside it, which is itself the clue that the meter, not the resistors, is the problem.
2. The student needs a trustworthy measurement of the divider's behavior. Propose two fixes and predict the reading each would give.
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Fix 1 - lower the circuit impedance. Replace the 10 MΩ resistors with 10 kΩ, 1% units. The same meter now puts 10 MΩ in parallel with 10 kΩ: R_p = (10 kΩ × 10 MΩ)/(10 MΩ + 10 kΩ) ≈ 9.99 kΩ, a loading error of about 0.01%. The reading becomes 9.00 × 9.99/(19.99) = 4.50 V. This is the standard remedy: keep the meter's input resistance at least 100 times the resistance across which you measure.
Fix 2 - use a better meter. With a 100 MΩ input resistance the parallel combination is 9.09 MΩ, giving 9.00 × 9.09/19.09 = 4.29 V - still 4.7% low. To reach 4.50 V within 1% the meter would need roughly 1 GΩ input resistance, which is why high-impedance probes exist for this kind of measurement.
A third option, worth naming: measure the battery voltage and one resistor's value with an ohmmeter (with power removed), then compute the divider ratio rather than measuring it under load. That avoids loading altogether, though it assumes the resistance is constant with voltage.
3. Suppose the student instead measured the 9.00 V battery directly with the same meter and got 9.00 V. Does that contradict the loading explanation? What does it tell you about the battery?
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It does not contradict the explanation. Measuring the battery puts the 10 MΩ meter directly across the source, drawing 9.00 V / 10 MΩ = 0.90 µA. That current flows through the battery's internal resistance; even a mediocre 9 V battery with, say, 10 Ω internal resistance drops only 9 µV, which is invisible at the meter's resolution. The reading is accurate because the source impedance is tiny compared with the meter's 10 MΩ, not because the meter is ideal.
The contrast is instructive: the same meter is perfectly adequate across a low-impedance source and grossly inadequate across a 10 MΩ divider. Whether a meter loads a circuit depends on the ratio of its input resistance to the Thevenin resistance of the thing being measured - roughly R_meter/(R_meter + R_Thevenin) as a correction factor. Here the divider's Thevenin resistance is 5 MΩ against the meter's 10 MΩ, a 33% error; the battery's is a few ohms, a 10⁻⁶ error.
A useful diagnostic habit: if a reading disagrees with prediction by far more than component tolerance, suspect the instrument's impedance before suspecting the components.