Q.Three resistances 2 ohm, 3 ohm and 4 ohm are connected in parallel. The ratio of currents passing through them when a potential difference is applied across its ends will be

Wheatstone Bridge Symmetry — Why the Formula Holds

The Wheatstone bridge is a circuit used to measure an unknown resistance by balancing two legs of a bridge. The key result is:

When the bridge is balanced, no current flows through the galvanometer, and:

$\frac{R_1}{R_2} = \frac{R_3}{R_4}$

Let's understand why this is true — not just memorize it.

1. The Circuit Setup

A Wheatstone bridge has four resistors arranged in a diamond:

• $R_1$ and $R_2$ in series on the left branch
• $R_3$ and $R_4$ in series on the right branch
• A galvanometer (sensitive current detector) connects the midpoints of the two branches
• A battery connects across the top and bottom

        A
       / \
      /   \
     R1    R3
     |     |
     G-----|   (G = galvanometer)
     |     |
     R2    R4
      \   /
       \ /
        B

2. The Condition for Balance — "No Current Through G"

Balance means the galvanometer shows zero deflection — no current flows through it. This implies points C (between $R_1$ and $R_2$) and D (between $R_3$ and $R_4$) are at the same electric potential. If $V_C = V_D$, no potential difference exists across the galvanometer, so no current flows.

3. Deriving the Ratio — Step by Step

Step 1: Branch currents

With no current through G, the left branch carries a single current $I_1$ and the right branch a single current $I_2$.

Step 2: Equal potentials at the midpoints

From A (common top) to the midpoints, since $V_C = V_D$:

From the midpoints to B (common bottom), since $V_C = V_D$:

Step 3: Divide (1) by (2)

The currents $I_1$ and $I_2$ cancel: …