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 increase in potential difference across the 200 ohm resistor as a result of connecting the voltmeter across it?​

Answer 1

Answer:

The voltage V

1

​

in the circuit before connecting the voltmeter is

V

1

​

=E−ir

V

1

​

=50−

220

50

​

×20

V

1

​

=50−4.6=45.4V

When the voltmeter is connected in the circuit as shown in figure, the voltage V

2

​

is

V

2

​

=50−

1800

50

​

×200

V

2

​

=44.4V

Now, percentage change in voltages is =

V

1

​

V

1

​

−V

2

​

​

×100=2.2%