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Derive Null Condition for Wheatstone Bridge

Question

Derive Null Condition for Wheatstone Bridge

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Solution

The Wheatstone Bridge is a circuit used for measuring the unknown resistance by balancing two legs of a bridge circuit, one leg of which includes the unknown component. The Null condition for a Wheatstone Bridge is when the bridge is balanced, i.e., no current flows through the galvanometer and the potential difference across it is zero.

Here are the steps to derive the Null condition for a Wheatstone Bridge:

  1. Consider a Wheatstone Bridge circuit with four resistances R1, R2, R3, and R4. Let's assume that the bridge is balanced, i.e., no current is flowing through the galvanometer G.

  2. According to Kirchhoff's second law, the sum of potential differences in any closed loop or mesh in a network is equal to the sum of the emf's in that loop. Applying this law to the first loop of the Wheatstone Bridge (R1, R2, and the battery), we get:

    V1 = I1 * R1 = I2 * R2

    where V1 is the potential difference across the battery, I1 is the current flowing through R1, and I2 is the current flowing through R2.

  3. Similarly, applying Kirchhoff's second law to the second loop (R3, R4, and the battery), we get:

    V2 = I3 * R3 = I4 * R4

    where V2 is the potential difference across the battery, I3 is the current flowing through R3, and I4 is the current flowing through R4.

  4. Since the bridge is balanced, the potential difference across the battery is the same for both loops, i.e., V1 = V2. Therefore, we can equate the two equations from steps 2 and 3 to get:

    I1 * R1 = I2 * R2 = I3 * R3 = I4 * R4

  5. This equation can be rearranged to give the Null condition for a Wheatstone Bridge:

    R1/R2 = R3/R4

This means that for the Wheatstone Bridge to be balanced (Null condition), the ratio of the resistances in the first loop must be equal to the ratio of the resistances in the second loop.

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