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 _________ $\Omega$.
Show Hint
For circuits that define their own equivalent resistance as an unknown $X$, you will usually end up with a quadratic equation in $X$.
Step 1: Understanding the Concept:
The circuit consists of resistors in a bridge-like configuration. We calculate the equivalent resistance by identifying series and parallel combinations between points A and B.
Step 2: Key Formula or Approach:
Based on the diagram, the equivalent resistance \( R_{eq} \) is found by taking the parallel combination of resistor \( R \) with the series branch containing \( 2R \) and \( X \):
\[ R_{eq} = \frac{(2R + X) \cdot R}{(2R + X) + R} \]
Step 3: Detailed Explanation:
Given that \( R_{eq} = X \):
\[ X = \frac{(2R + X)R}{3R + X} \]
\[ X(3R + X) = 2R^2 + RX \]
\[ 3RX + X^2 = 2R^2 + RX \]
\[ X^2 + 2RX - 2R^2 = 0 \]
Using the quadratic formula \( X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ X = \frac{-2R \pm \sqrt{(2R)^2 - 4(1)(-2R^2)}}{2} \]
\[ X = \frac{-2R \pm \sqrt{4R^2 + 8R^2}}{2} = \frac{-2R \pm \sqrt{12R^2}}{2} \]
\[ X = \frac{-2R \pm 2R\sqrt{3}}{2} = -R \pm R\sqrt{3} \]
Since resistance cannot be negative, we take the positive root:
\[ X = (\sqrt{3} - 1)R \]
Step 4: Final Answer:
The value of $X$ is $(\sqrt{3} - 1)R$.
