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Question

A voltmeter having a resistance of 1800 Ω is employed to measure the potential difference across 200Ω resistance, which is connected to dc power supply of 50 V and internal resistance 20Ω. What is the approximate percentage change in the potential difference across 200Ω resistance as a result of connecting the voltmeter across it?
119390_f6e066f8565d4dd6bddf3fade2ed7b82.png

A
2.2%
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B
5%
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C
10%
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D
20%
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Solution

The correct option is B 2.2%
The voltage V1 in the circuit before connecting the voltmeter is
V1=Eir
V1=5050220×20
V1=504.6=45.4V
When the voltmeter is connected in the circuit as shown in figure, the voltage V2 is
V2=50501800×200
V2=44.4V
Now, percentage change in voltages is =V1V2V1×100=2.2%

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