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

(a)
`V_0 = (20)/(20 + 10) xx 100 = (200)/(3) V`
Reading of voltmeter `= (200)/(3) V`
(b)
`p.d` across `2 Omega : 40 V`
Current through it `i= (40)/(20) = 2 A`
`i_1 = (40)/(60) = (2)/(3) A`
`i= i_1 + i_V`
`2 = (2)/(3) + i_V rArr i_V rArr i_V = (4)/(3) A`
Voltmeter :
`V_V = i_V R_V`
`40 = (40)/(3) R_V`
`R_V = 30 Omega`
Resistance of voltmeter `= 30 Omega`
( c)
`V_X - V_Y = (2R)/(R + 2R + 2R) xx E = (2E)/(5) = 0.4 E`
If voltmeter is ideal i.e. `R_V = oo`

`V_X - V_Y = (3R)/(R + 2R + 3R) xx E = ( E)/(2) = 0.5 E`
`% error = ((0.4 E)/(0.5 E) - 1) xx 100`
=`-20 %`
i.e. `20 %` smaller.