Question

In the following circuit diagram, when the \(3\Omega\) resistor is removed, the equivalent resistance of the network: 
 

• A. Increases
• B. Decreases
• C. Remains the same
• D. None of these

Hint:

In a balanced Wheatstone bridge, removing a resistor from one arm does not affect the total equivalent resistance, as no current flows through that branch when the bridge is balanced.

Answer 1

The Correct Option is C

Step 1: {Identifying Wheatstone Bridge} 
The given network forms a balanced Wheatstone bridge. In a Wheatstone bridge, four resistors are arranged in a diamond shape, with two resistors on each arm, and a galvanometer connected across the middle of the bridge. In this case, when the \( 3 \, \Omega \) resistor in the BD arm is removed, the bridge remains balanced. 
Key Concept: A Wheatstone bridge is balanced when the ratio of resistances in one pair of opposite arms is equal to the ratio of resistances in the other pair. The condition for balance in the Wheatstone bridge is given by: \[ \frac{R_1}{R_2} = \frac{R_3}{R_4} \] Where \( R_1, R_2, R_3, \) and \( R_4 \) are the resistances in the four arms of the bridge. When the \( 3 \, \Omega \) resistor is removed from the BD arm, it does not affect the total resistance because the bridge is balanced and the current flowing through the galvanometer is zero, meaning no current flows through the branch containing the \( 3 \, \Omega \) resistor. Hence, the equivalent resistance of the bridge does not change. 
Step 2: {Finding Equivalent Resistance} 
Since the bridge remains balanced, the equivalent resistance of the network remains unchanged. For a balanced Wheatstone bridge, the total equivalent resistance across the bridge is determined by the resistances in the remaining arms. Let’s denote the resistances of the other arms as \( R_1, R_2, R_3, \) and \( R_4 \). When the bridge is balanced, we know the equivalent resistance for the two parallel arms (AC and BD) can be calculated as: \[ R_{{eq}} = \frac{R_1 R_2}{R_1 + R_2} \] This is true for both the upper and lower parts of the Wheatstone bridge, and they contribute equally to the total equivalent resistance. Since the \( 3 \, \Omega \) resistor is removed from the BD arm, it does not change the equivalent resistance of the network because the Wheatstone bridge was balanced and no current flows through the removed resistor. The equivalent resistance remains as calculated previously. 
Step 3: {Conclusion} 
Thus, the total equivalent resistance of the bridge remains unchanged when the \( 3 \, \Omega \) resistor is removed.

Answer 2

Step 1: Understand the circuit layout
The original circuit includes the following resistors:
- \( 1\Omega \) between B and C
- \( 2\Omega \) between A and B
- \( 3\Omega \) between B and D
- \( 2\Omega \) between D and C
- \( 4\Omega \) between A and D

Step 2: Remove the \( 3\Omega \) resistor
According to the question, the \( 3\Omega \) resistor between B and D is removed.
So now, the direct connection between B and D is eliminated.

Step 3: Analyze current flow
Note that point B and D were connected via a resistor of \( 3\Omega \), but removing it will disconnect direct current flow between those two points.
The current cannot flow from B to D without this resistor, which means paths through 1Ω (from B to C) and 2Ω (from D to C) are now unaffected.

Step 4: Observe the unchanged current paths
Even with the \( 3\Omega \) resistor removed, current can still pass from A to B to C, and separately from A to D to C.
Since the paths from A–B–C and A–D–C remain intact and unaffected by the removal of the B–D resistor, the total or equivalent resistance of the circuit remains the same.

Final Answer: Remains the same