Two known resistances of R Omega and 2R Ome | vedclass

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(A) From the circuit diagram, the resistance $2R$ and $X$ are in series. Their equivalent resistance is $R_{s} = 2R + X$.This combination is in parall

Vedclass · Accepted Answer

$(\sqrt{3}-1)R$

Vedclass · Answer

(A) From the circuit diagram, the resistance $2R$ and $X$ are in series. Their equivalent resistance is $R_{s} = 2R + X$.This combination is in parallel with the resistance $R$. The equivalent resistance $R_{eq}$ between points $A$ and $B$ is given by:$R_{eq} = \frac{R \cdot (2R + X)}{R + (2R + X)}$Given that $R_{eq} = X$, we have:$X = \frac{R(2R + X)}{3R + X}$$X(3R + X) = 2R^{2} + RX$$3RX + X^{2} = 2R^{2} + RX$$X^{2} + 2RX - 2R^{2} = 0$Using the quadratic formula $X = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$:$X = \frac{-2R \pm \sqrt{(2R)^{2} - 4(1)(-2R^{2})}}{2(1)}$$X = \frac{-2R \pm \sqrt{4R^{2} + 8R^{2}}}{2}$$X = \frac{-2R \pm \sqrt{12R^{2}}}{2} = \frac{-2R \pm 2R\sqrt{3}}{2}$$X = -R \pm R\sqrt{3}$Since resistance must be positive, we take the positive root:$X = (\sqrt{3} - 1)R$

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

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