Difficulty: Easy
Correct Answer: The meter resistance is at least 10 times greater than the resistance being measured across
Explanation:
Introduction / Context:
Real voltmeters are not ideal; they have finite input resistance R_m that loads the circuit and can pull the measured node down. Engineers use a simple criterion to judge whether this loading is negligible: the 10× rule of thumb.
Given Data / Assumptions:
Concept / Approach:
When you connect a meter across a resistor R_x, the effective resistance becomes R_eff = (R_x * R_m) / (R_x + R_m). If R_m ≫ R_x, then R_eff ≈ R_x and the circuit is barely disturbed. A common practical threshold is R_m ≥ 10 * R_x, which keeps loading error typically under about 10% and often much less in practice, especially with modern DMMs (10 MΩ inputs).
Step-by-Step Solution:
Verification / Alternative check:
Example: R_x = 100 kΩ; DMM input R_m = 10 MΩ (100×). The equivalent is ~99.01 kΩ, producing <1% error in most divider readings — negligible for many applications.
Why Other Options Are Wrong:
Common Pitfalls:
Confusing meter range with input resistance; some analog meters have low input resistance on low-voltage ranges, causing significant loading errors on high-impedance circuits.
Final Answer:
Use the 10× rule — R_m should be at least 10 times R_x to ignore loading.
Discussion & Comments