1.

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

Answer»

`2.2%`
`5%`
`10%`
`20%`

Solution :`V_(1) = E - ir = 50 - (50)/(220) xx 20`
`= 50 - 4.6 = 45.4 V`
Now, `V_(2) = 50 - (50)/(180) xx 20 = 44.4 V`
percentage change `= (V_(1) - V_(2))/(V_(1)) xx 100`
`= 2.27` (also see the question)


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