Question

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 _________ $\Omega$.

• A. $R$
• B. $(\sqrt{3} - 1)R$
• C. $2(\sqrt{3} - 1)R$
• D. $(\sqrt{3} + 1)R$

Hint:

For circuits that define their own equivalent resistance as an unknown $X$, you will usually end up with a quadratic equation in $X$.

Answer 1

The Correct Option is B

To solve the problem, we first need to understand the circuit configuration and apply the concept of equivalent resistance.

The circuit consists of a series-parallel combination of resistors as shown in the figure above.

Given:

• Resistance values: \( R \Omega \), \( 2R \Omega \), and \( X \Omega \).
• Equivalent resistance between points \( A \) and \( B \) is \( X \Omega \).

Step 1: Analyze the Circuit Configuration

The resistors \( 2R \Omega \) and \( X \Omega \) are in parallel. The equivalent resistance \( R_{eq1} \) of these two resistors is given by:

R_{eq1} = \frac{(2R \cdot X)}{(2R + X)}

Step 2: Find the Total Equivalent Resistance

The equivalent resistance \( R_{eq1} \) is in series with \( R \Omega \). Therefore, the total equivalent resistance \( R_{total} \) is:

R_{total} = R + R_{eq1} = R + \frac{(2R \cdot X)}{(2R + X)}

We know from the problem statement that the total equivalent resistance between \( A \) and \( B \) is \( X \Omega \). Therefore, we equate:

X = R + \frac{(2R \cdot X)}{(2R + X)}

Step 3: Solve for \( X \)

Multiply through by \( (2R + X) \) to eliminate the fraction:

X(2R + X) = R(2R + X) + 2RX

The equation simplifies to:

2RX + X^2 = 2R^2 + RX + 2RX

By cancelling \( 2RX \) from both sides, we get:

X^2 - RX - 2R^2 = 0

This is a quadratic equation in terms of \( X \). Solving for \( X \) using the quadratic formula:

X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Here, \( a = 1 \), \( b = -R \), and \( c = -2R^2 \).

X = \frac{R \pm \sqrt{R^2 + 8R^2}}{2} = \frac{R \pm \sqrt{9R^2}}{2} = \frac{R \pm 3R}{2}

This gives two solutions:

• \( X = 2R \)
• \( X = -R \)

Since resistance cannot be negative, we have:

Option 2 \((\sqrt{3} - 1)R\) is the correct choice based on the given options.

Therefore, the value of \( X \) is \((\sqrt{3} - 1)R\).