Analog meter loading: For a 20,000 Ω/volt voltmeter set to its 5 V range, determine the internal resistance presented to the circuit.

Difficulty: Easy

Correct Answer: 100,000 Ω

Explanation:


Introduction / Context:
The Ω/volt rating indicates how much resistance a voltmeter exhibits per volt of full-scale range. This parameter matters because finite internal resistance can load the circuit and alter the measured voltage, especially in high-resistance networks.


Given Data / Assumptions:

  • Sensitivity: 20,000 Ω/V.
  • Range: 5 V.
  • Use R_internal = (Ω/V) × (range).


Concept / Approach:

On each higher voltage range, additional series resistance is introduced so that the meter movement sees the correct full-scale current. Therefore the presented resistance scales linearly with range setting, following the product rule above.


Step-by-Step Solution:

Compute: R_internal = 20,000 Ω/V * 5 V.Result: R_internal = 100,000 Ω (100 kΩ).Thus the meter loads the node with a 100 kΩ equivalent resistance on the 5 V range.


Verification / Alternative check:

On a 10 V range, the same instrument would show 200 kΩ; on 1 V, 20 kΩ. The proportionality confirms the logic.


Why Other Options Are Wrong:

20,000 Ω corresponds to 1 V range, not 5 V. 200,000 Ω would be the 10 V range value. 1,000,000 Ω implies 200,000 Ω/V sensitivity, not 20,000 Ω/V.


Common Pitfalls:

Forgetting to multiply by the selected range; confusing Ω/V with kΩ/V; overlooking how meter loading changes with range.


Final Answer:

100,000 Ω

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