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Wheatstone Bridge and Balance Condition

A Wheatstone bridge is a four-resistor network used to compare resistances. It is balanced when no current flows through the galvanometer branch, giving the condition P/Q = R/S.

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Student-friendly explanation

In a Wheatstone bridge, four resistors form two potential-divider arms, and a galvanometer connects the middle points. At balance, the two middle points are at the same potential, so the galvanometer shows null deflection and carries no current. The balance condition then depends only on the ratio of resistances in the two arms, not on the galvanometer resistance. This principle is used in resistance measurement and in the metre bridge.

How to write this in exams

  1. 1

    Start with the exact idea

    A Wheatstone bridge is a four-resistor network used to compare resistances. It is balanced when no current flows through the galvanometer branch, giving the condition P/Q = R/S.

  2. 2

    Then show how to use it

    Label the four arms correctly as P, Q, R, and S. Confirm that the bridge is balanced or the galvanometer current is zero. Write P/Q = R/S or PS = QR. Substitute known resistances with ohm units. Solve for the unknown. State that galvanometer current is zero at balance.

  3. 3

    Add one concrete example

    If P = 2 ohm, Q = 4 ohm, and R = 3 ohm, the bridge is balanced when 2/4 = 3/S, so S = 6 ohm.

  4. 4

    Avoid this incomplete answer

    Using the balance formula when the bridge is not balanced is wrong because current then flows through the galvanometer branch and simple ratio equality does not apply.

Definition

A Wheatstone bridge is a four-resistor network used to compare resistances. It is balanced when no current flows through the galvanometer branch, giving the condition P/Q = R/S.

Example

If P = 2 ohm, Q = 4 ohm, and R = 3 ohm, the bridge is balanced when 2/4 = 3/S, so S = 6 ohm.

Rule to remember

Balance condition: P/Q = R/S, where P and Q are resistances in one arm and R and S are corresponding resistances in the other arm, all measured in ohm. Equivalent form: PS = QR. This condition applies only at null deflection, when galvanometer current is zero.

Memory hook

Balanced bridge means equal middle potentials, so the galvanometer has nothing to detect.

Examples and method

Worked example

A balanced Wheatstone bridge has P = 10 ohm, Q = 20 ohm, and R = 15 ohm. Using P/Q = R/S, 10/20 = 15/S. Therefore S = 30 ohm. Since the bridge is balanced, the galvanometer current is zero and the unknown resistance is 30 ohm.

Method to apply

Label the four arms correctly as P, Q, R, and S. Confirm that the bridge is balanced or the galvanometer current is zero. Write P/Q = R/S or PS = QR. Substitute known resistances with ohm units. Solve for the unknown. State that galvanometer current is zero at balance.

Diagram support

Draw a diamond-shaped or rectangular bridge with resistors P, Q, R, and S in four arms, a galvanometer between the two middle junctions, and a cell across the other pair of junctions. Mark the galvanometer current as zero at balance.

How CBSE asks it

Questions ask for unknown resistance at balance, explanation of null deflection, relation with metre bridge, or why the galvanometer branch can be ignored only under balance condition.

Avoid common mistakes

Common confusion

Students sometimes use P/R = Q/S without checking the actual arm positions. The ratio must compare the two resistors in one arm with the corresponding two resistors in the other arm as labelled in the circuit.

Common wrong answer

Using the balance formula when the bridge is not balanced is wrong because current then flows through the galvanometer branch and simple ratio equality does not apply.

Exam tip

At balance, treat the galvanometer branch as carrying zero current. This allows the bridge network to be simplified using series relations in each arm.

Quick check

Why does no current flow through the galvanometer in a balanced Wheatstone bridge?

No current flows through the galvanometer because the two points connected to it are at the same potential. With zero potential difference across the galvanometer branch, the galvanometer shows null deflection.

Answer writing and exam use

1-mark answer

A Wheatstone bridge is a four-resistor network used to compare resistances. It is balanced when no current flows through the galvanometer branch, giving the condition P/Q = R/S.

2-mark answer

A Wheatstone bridge is a four-resistor network used to compare resistances. It is balanced when no current flows through the galvanometer branch, giving the condition P/Q = R/S. Balance condition: P/Q = R/S, where P and Q are resistances in one arm and R and S are corresponding resistances in the other arm, all measured in ohm. Equivalent form: PS = QR. This condition applies only at null deflection, when galvanometer current is zero. If P = 2 ohm, Q = 4 ohm, and R = 3 ohm, the bridge is balanced when 2/4 = 3/S, so S = 6 ohm.

3-mark answer

In a Wheatstone bridge, four resistors form two potential-divider arms, and a galvanometer connects the middle points. At balance, the two middle points are at the same potential, so the galvanometer shows null deflection and carries no current. The balance condition then depends only on the ratio of resistances in the two arms, not on the galvanometer resistance. This principle is used in resistance measurement and in the metre bridge. Balance condition: P/Q = R/S, where P and Q are resistances in one arm and R and S are corresponding resistances in the other arm, all measured in ohm. Equivalent form: PS = QR. This condition applies only at null deflection, when galvanometer current is zero. A balanced Wheatstone bridge has P = 10 ohm, Q = 20 ohm, and R = 15 ohm. Using P/Q = R/S, 10/20 = 15/S. Therefore S = 30 ohm. Since the bridge is balanced, the galvanometer current is zero and the unknown resistance is 30 ohm. Questions ask for unknown resistance at balance, explanation of null deflection, relation with metre bridge, or why the galvanometer branch can be ignored only under balance condition. Using the balance formula when the bridge is not balanced is wrong because current then flows through the galvanometer branch and simple ratio equality does not apply.
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