Question

In a Wheatstone's bridge all the four arms have equal resistance R. If the resistance of the galvanometer arm is also R,the equivalent resistance of the combination as seen by the battery is:

• A. \(\frac{R}{4}\)
• B. \(\frac{R}{2}\)
• C. R
• D. 2R

Answer 1

The Correct Option is C

The given problem involves a Wheatstone's bridge where all the arms, including the galvanometer arm, have equal resistance \( R \). We need to determine the equivalent resistance of the bridge as viewed from the battery's terminal.

Let's approach this step-by-step:

1. A Wheatstone's bridge is a special type of circuit used to measure unknown resistances. It consists of four resistances arranged in a quadrilateral, with a galvanometer between two opposite corners.

2. When the bridge is balanced, the ratio of the resistances in one diagonal is equal to the ratio in the other diagonal:

   \(\frac{R_1}{R_2} = \frac{R_3}{R_4}\)

3. In this question, since all four arms have equal resistance, the bridge is inherently balanced:

   \(\frac{R}{R} = \frac{R}{R}\)

4. For a balanced Wheatstone's bridge, no current flows through the galvanometer. Therefore, it can be removed from the circuit, simplifying the calculation of equivalent resistance.

5. The remaining resistances form two parallel branches, each with series resistors. In each branch, the two resistors in series add up to:

   R + R = 2R

6. These two branches are in parallel, so the equivalent resistance R_{\text{eq}} of the entire Wheatstone bridge is given by the formula for parallel resistors:

   \(\frac{1}{R_{\text{eq}}} = \frac{1}{2R} + \frac{1}{2R}\)

7. Simplifying the above equation, we get:

   \(\frac{1}{R_{\text{eq}}} = \frac{1}{R}\)

   Therefore, R_{\text{eq}} = R.

Thus, the equivalent resistance of the combination as seen by the battery is R.