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

Applying kirchhoff's law in loop `JMGDJ`, we get

`-(I_(3))/(2) xx 400 + (I_(1) + I_(2) - I_(3)) 200 + (I_(2) - I_(3)) 100 = 0`
or `- 500 I_(3) + 200 I_(1) + 300 I_(2) = 0` or `2 I_(1) + 3I_(2) - 5I_(3) = 0` (i)
Applying kirchhoff's law in `CDEFBC`, we get
`- 100I_(2) - 100(I_(2) - I_(3)) + 100I_(1) = 0`
or `I_(1) - 2I_(2) + I_(3) = 0` (ii)
Mulplying Eq. (i) by `2` and subracting from Eq. (i), we have
`2I_(1) + 3I_(2) - 5I_(3) - 2I_(1) + 4I_(2) - 2I_(3) = 0` or `I_(2) = I_(3)` (iii)
Applying kirchhoff's law in `ABEFGHA`, we get
`- 3I_(1) - 2I_(2) + 2I_(3) + 0.1 = 0` (iv)
Multiplying Eq.(ii) by `3` and adding it to Eq.(iv), we get
`3I_(1) - 6I_(2) + 3I_(3) - 3I_(1) - 2I_(2) + 2I_(3) + 0.1 = 0`
or `-8I_(2) + 5I_(3) + 0.1 = 0` or `-8I_(3) + 5I_(3) + 0.1 = 0`
or `3I_(3) = 0.1` or `I_(3) = (0.1)/(3)`
Therefore, potential difference across `JM` is
`(0.1//3)/(2) xx 400 = (0.1)/(2 xx 3) xx 400`
`= (20)/(3) = 6.67 V`
Alternative method
We can redraw the circuit as shows in Fig. `6.26`. The equivalent resistance between `G` and `D` is
`R_(GD) = (400 xx 400)/(400 + 400) = 200 Omega ("Since" (R_(GE))/(R_(GD)) = (R_(EB))/(R_(DB))) `
Therefore, it is a case of a balanced wheatstone bridge. The equivalent resistance across `G` and `B` is
`R_(GB) = (300 xx 300)/(300 + 300) = 150 Omega`
Therefore, current is
`I = (V)/(R_(GB)) = (10)/(150) = (1)/(15) A`

Since `RGEB = RGDB`, the current is divided at `G` into two equal
parts. The current `I//2` is further divided into two equal parts at
`M`. Therefore, the potential difference across the voltmeter is
`(1)/(4) xx 400 = (1)/(15) xx 100 = (20)/(3) V`