An electrical circuit is shown in figure. Calculate the potential difference across the resistor of `400 Omega` as will be measured by the voltmeter Vof resistance `400 Omega` either by applying Kirchhoff's rules or otherwise.

Explanation

The given circuit actually forms a balanced Wheatstone’s bridge (including the voltmeter) as shown below
`R_v=400 Omega`, Here we see that `P/Q=R/S`
Therefore, resistance between can be ignored and equivalent simple circuit can be drawn as follows. The voltmeter will read the potential difference across resistance Q
Currents `j_1=j_2=10/(100+20)=1/30A`

`therefore` Potential difference across voltmeter , `=Qj_1=(200)(1/30)V=20/3V`.
Therefore, reading of voltmeter will be `20/3V`