Question

The equivalent resistance between the terminal points \(X\) and \(Y\) in the given circuit is

• A. \(15~\Omega\)
• B. \(\sqrt{3}~\Omega\)
• C. \(40~\Omega\)
• D. \(30~\Omega\)

Hint:

In bridge-network problems, always check the balance condition first: \[ \frac{R_1}{R_2}=\frac{R_3}{R_4}. \] If the bridge is balanced, the central branch carries no current and can be removed, making the circuit much easier to solve.

Answer 1

The Correct Option is D

Concept: The given network is a bridge network. Before applying series-parallel reduction, we should check whether the bridge is balanced. For a balanced bridge, \[ \frac{R_1}{R_2} = \frac{R_3}{R_4}. \] In that case, no current flows through the bridge branch and it can be removed from the circuit for equivalent resistance calculation.

Step 1: Identify the bridge arms.
The four outer arms of the network are: \[ 15\Omega,\quad 30\Omega,\quad 15\Omega,\quad 30\Omega. \] Checking the balance condition, \[ \frac{15}{30} = \frac{15}{30} = \frac12. \] Since both ratios are equal, the bridge is balanced. Therefore, no current flows through the middle \(30\Omega\) resistor. Hence, the middle branch can be ignored.

Step 2: Reduce the upper path between \(X\) and \(Y\).
The upper path consists of \[ 30\Omega + 15\Omega + 15\Omega. \] Thus, \[ R_{\text{upper}} = 60\Omega. \]

Step 3: Reduce the lower path between \(X\) and \(Y\).
The lower path consists of \[ 15\Omega + 15\Omega + 30\Omega. \] Therefore, \[ R_{\text{lower}} = 60\Omega. \]

Step 4: Combine the two parallel branches.
The circuit now becomes two \(60\Omega\) resistors connected in parallel. Hence, \[ R_{XY} = \frac{60\times60}{60+60}. \] \[ = \frac{3600}{120}. \] \[ = 30\Omega. \]

Step 5: Write the final answer.
Therefore, the equivalent resistance between terminals \(X\) and \(Y\) is \[ \boxed{30~\Omega}. \]