Learning objective: Calculate percentage-of-reading error from full-scale specifications and understand the loading caused by finite meter resistance.
Understand the idea
An electrical meter can report an inaccurate value for two distinct reasons. Its internal measurement tolerance creates uncertainty, while its finite input resistance changes the circuit being measured. GATE instrumentation problems often test both effects, so identify whether the question asks about meter accuracy or loading error.
Worked example: given data
- A voltmeter is specified as class 1% of full scale on its 200 V range.
- It displays 50 V. Calculate the maximum stated absolute error and the corresponding percentage of the indicated reading.
- Separately, a 10 V source with two equal 10 kΩ series resistors is measured across the lower resistor by a 10 kΩ input-resistance voltmeter. Find the loading error.
Detailed solution
Apply the full-scale accuracy specification
The absolute accuracy limit is ±1% of 200 V = ±2 V. The tolerance is not ±1% of the indicated reading unless the specification explicitly states that.
Express error relative to 50 V
At 50 V indication, the stated ±2 V corresponds to (2/50) × 100 = ±4% of reading. This shows why a lower suitable measurement range can improve relative accuracy.
Find the unloaded divider voltage
With two equal 10 kΩ resistors across 10 V, the ideal voltage across the lower resistor is 10 × 10/(10 + 10) = 5 V.
Account for meter input resistance
The 10 kΩ voltmeter lies in parallel with the lower 10 kΩ resistor; their equivalent is 5 kΩ. The loaded divider now has 10 kΩ above and 5 kΩ below, giving Vmeasured = 10 × 5/(10 + 5) = 3.333 V.
Calculate loading error
The loading error is (3.333 − 5)/5 × 100 ≈ −33.3%. This large error is caused by meter loading, not the 1%-of-full-scale tolerance.
What the answer means
Improving a voltmeter's stated accuracy class is not enough if its input resistance is comparable with the circuit resistance. High-input-impedance instruments minimize loading. Check the operating range and measurement frequency before choosing equipment for an actual circuit.
Three mistakes to avoid
- Mistaking percentage of full scale for percentage of reading.
- Adding the voltmeter resistance in series instead of in parallel with the measured element.
- Combining loading error and instrument uncertainty without clarifying the assumptions.
Try a related question
On a 100 V range with the same 1%-of-full-scale accuracy, what is the absolute tolerance and percentage of reading at 50 V?
Show the worked-answer check
The tolerance becomes ±1 V, corresponding to ±2% of the 50 V reading. This statement concerns accuracy; loading depends on the meter input resistance.
How to revise this concept
Compare analog instrument classes with digital multimeter specifications such as ±(% of reading + counts). First attempt the example on paper without opening the answer, then compare the intermediate steps. Explain each assumption before memorising a formula.
Content note: This is a newly prepared practice example, not an official GATE previous-year question. It was drafted with AI assistance for this educational website. Students should verify calculations and exam requirements against standard textbooks and official sources; educator review is recommended before publication.