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

### Related Formula For parallel combination: R_texteq = fracR_1 R_2R_1 + R_2$R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2}$ For series combination: R_texteq = R_1 + R_2$R_{\text{eq}} = R_1 + R_2$ ### Core Logic Based on the circuit topology, the 2R$2R$ and X$X$ resistors are in series. This combination is in parallel with the R$R$ resistor. The total equivalent resistance across A and B is given as X$X$. R_texteq = frac(2R + X) cdot (R)(2R + X) + R$$R_{\text{eq}} = \frac{(2R + X) \cdot (R)}{(2R + X) + R}$$ ### Step 1: Setting up the Quadratic Set the equivalent resistance equal to X$X$: X = frac(2R + X)R3R + X$$X = \frac{(2R + X)R}{3R + X}$$ Cross-multiplying yields: X(3R + X) = 2R^2 + XR$$X(3R + X) = 2R^2 + XR$$ 3RX + X^2 = 2R^2 + XR$$3RX + X^2 = 2R^2 + XR$$ X^2 + 2RX - 2R^2 = 0$$X^2 + 2RX - 2R^2 = 0$$ ### Step 2: Solving for X Using the quadratic formula to solve for X$X$: X = frac-2R pm sqrt(2R)^2 - 4(1)(-2R^2)2$$X = \frac{-2R \pm \sqrt{(2R)^2 - 4(1)(-2R^2)}}{2}$$ X = frac-2R pm sqrt4R^2 + 8R^22 = frac-2R pm sqrt12R^22$$X = \frac{-2R \pm \sqrt{4R^2 + 8R^2}}{2} = \frac{-2R \pm \sqrt{12R^2}}{2}$$ X = frac-2R pm 2Rsqrt32 = -R pm Rsqrt3$$X = \frac{-2R \pm 2R\sqrt{3}}{2} = -R \pm R\sqrt{3}$$ Since resistance must be positive: X = (sqrt3 - 1)R$$X = (\sqrt{3} - 1)R$$ ### Pattern Recognition Self-referential equivalent resistance problems always collapse into a quadratic equation x^2 + ax + b = 0$x^2 + ax + b = 0$. Solve strictly for the positive root. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity