Question

The net resistance between the points A and B in the circuit given below is:

• A. \(\frac{8}{3}R\)
• B. \(\frac{8}{5}R\)
• C. \(R\)
• D. \(\frac{5}{3}R\)

Hint:

In a balanced Wheatstone bridge, no current flows through the middle resistor, so it can be removed while finding equivalent resistance.

Answer 1

The Correct Option is A

Step 1: Identify the terminals.
The resistance is to be found between points \(A\) and \(B\). Point \(A\) lies on the left lower junction and point \(B\) lies on the inner junction.

Step 2: Observe the bridge network.
The resistors connected between the outer and inner junctions form a bridge-like circuit.

Step 3: Check the balance condition.
On the upper branch, the resistances connected to the top junction are:
\[ 4R \quad \text{and} \quad 4R \]
On the lower branch, the resistances connected to the right lower junction are:
\[ 2R \quad \text{and} \quad 2R \]
Since the ratios are equal, the bridge is balanced.
\[ \frac{4R}{4R} = \frac{2R}{2R} \]

Step 4: Remove the balanced bridge resistor.
The \(5R\) resistor has no current through it because both of its ends are at the same potential. Hence, it can be ignored.

Step 5: Simplify the remaining circuit.
Now, between \(A\) and \(B\), two branches are in parallel:
First branch:
\[ 4R + 4R = 8R \]
Second branch:
\[ 2R + 2R = 4R \]

Step 6: Find equivalent resistance.
\[ R_{eq} = \frac{8R \times 4R}{8R + 4R} \]
\[ R_{eq} = \frac{32R^2}{12R} \]
\[ R_{eq} = \frac{8}{3}R \]

Step 7: Final answer.
\[ \boxed{\frac{8}{3}R} \]