A voltmeter having a resistance of `1800 Omega` employed to measure the potential difference across a `200 Omega` resistor which is connected to the terminals of a dc power supply having an emf of 50 V and an internal resistance of `20 Omega`. What is the percentage decrease in the potential difference across the `200 Omega` resistor as a result of connecting the voltmeter across it?

Options

A. `2.2%`
B. `5%`
C. `10%`
D. `20%`

Answer

A

Explanation

`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)