An elctrical circuit is shown in Fig. Calculate the potential difference across the resistor of 400 ohm, as will be measured by the voltmeter V of resistance 400 ohm, either by applying Kirchhoff's rules or otherwise.
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Explanation

We can redraw the circuit as.

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))`
`:. ` It is a case of balanced wheatstone bridge.
The equivalent resistance across G and B is
`R_(GB) = (300 xx 300)/(300 + 300) = 150 Omega`
`:. Current I = V/(R_(GB)) = 10/150 = 1/15 Amp. `
NOTE : Since `R_(GEB) = R_(GDB)` the current is divided at G into
two equal parts (I/2). The current 1/2 further divides into
two equal parts at M.
Therefore, the potential difference across the voltmeter
` I/4 xx 400 = 1/15 xx 400 /4 = 20/3 volt = 6.67 V` .