Four resistances 10 Ω, 5 Ω, 7 Ω and 3 Ω are connected so that they form the sides of a rectangle AB, BC, CD and DA respectively. Another resistance of 10 Ω is connected across the diagonal AC. The equivalent resistance between A and B is
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$3\Omega$ resistor and $7\Omega$ resistor are in series.
Therefore resultant is $= 10 \Omega (7 + 3)$ .
This $10\Omega$ equivalent resistance is in parallel with resistance $(10\Omega)$ in arm AC.
$\therefore \frac{1}{R_1} = \frac{1}{10} + \frac{1}{10}$ $\Rightarrow R_1 = 5\Omega$ Now, $R_1$ is in series with resistor (5$\Omega$) in arm CB. $\therefore R_2 = 5 + 5 = 10\Omega$ Again $R_2$ is in parallel with resistance (10$\Omega$) in arm AB $\therefore \frac{1}{R} = \frac{1}{10} + \frac{1}{10}$ $\Rightarrow R = 5\Omega$
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