Two known resistances of R Ω and 2R Ω and one unknown resistance X Ω are connected in a circuit as shown in the figure. If the equivalent resistance between points A and B in the circuit is X Ω, then the value of X is __________ Ω.

Solution

Physics sir · 4 steps

The top branch holds X itself, so set the parallel combination equal to X and solve the quadratic.

1. Read the circuit. The top branch has 2R and X in series, which is 2R + X. This branch is in parallel with the bottom resistor R between A and B.

   X = (R (2R + X))/(R + (2R + X)) = (R (2R + X))/(3R + X)

2. Form the quadratic. Cross-multiply: 3RX + X² = 2R² + RX. This tidies up to X² + 2RX - 2R² = 0.
3. Solve it. X = (-2R ± √(4R² + 8R²))/2 = (-2R ± 2√3R)/2 = (-1 ± √3)R. A resistance cannot be negative, so take the plus sign.
4. So. X = (√3 - 1)R ≈ 0.732R, which is less than R, as any combination in parallel with R must be. The answer is (c).

The idea to keep. When the unknown sits inside the network and also equals the equivalent resistance, write R_(eq) = X, get a quadratic, and keep only the positive root.

Answer from the NTA final answer key. Solution by PrepWiser.