CGPET 2021 Solved Paper

Given, effective resistance between points A and D, $R_{AD} = 3\ \Omega$ The circuit is shown as Resistance of sides AC and CB are in series ∴ $R_1 = 3+3 = 6\ \Omega$.....(i) Resistance $R_1$ and R are in paralel ∴ $\frac{1}{R_2} = \frac{1}{R} + \frac{1}{R_1}$ ⇒ $R_2 = \frac{R \cdot R_1}{R+R_1}$ .....(ii) Resistance of sides AB and BD are series, $R_3=R_2+3$ ....(iii) Now, resistance $R_3$ and 3Ω are parallel ∴ $R_4=\frac{R_3 \times 3}{R_3+3}$.......(iv) Put the value from Eqs. (i), (ii) and (iii) in Eqs. (iv), we get $R_4=\frac{(R_2+3)3}{(R_2+3)+3} = \frac{3R_2+9}{R_2+6}$ $=\frac{3\left(\frac{R R_1}{R+R_1}\right)+9}{\left(\frac{R R_1}{R+R_1}\right)+6}$ ⇒ $3=\frac{\left(\frac{6R}{R}+6\right)+9}{\left(\frac{6R}{R}+6\right)+6} \quad (\text{Given, } R_4 = R_{AD} = 3\ \Omega)$ ⇒$R=6\ \Omega$