Question

Explain the principle of Wheatstone's bridge by Kirchhoff's law. In the given circuit, there is no deflection in the galvanometer \( G \). What is the current flowing through the cell?

Hint:

In a Wheatstone bridge, the bridge is balanced when the ratio of resistances in one leg is equal to the ratio of resistances in the other leg. Use Kirchhoff's law to solve for unknown values in the circuit.

Answer 1

The Wheatstone bridge is a circuit used to measure an unknown resistance by balancing two legs of a bridge circuit. The principle of the Wheatstone bridge states that when the bridge is balanced, the ratio of resistances in one leg equals the ratio of resistances in the other leg, which is derived from Kirchhoff's law. The balance condition of the Wheatstone bridge is given by: \[ \frac{R_1}{R_2} = \frac{R_3}{R_4} \] Where: - \( R_1, R_2, R_3, R_4 \) are the resistors in the four arms of the bridge,
- \( G \) is the galvanometer, which detects the balance condition by indicating zero deflection when the bridge is balanced.
In the given circuit: \[ R_1 = 3 \, \Omega, \, R_2 = 4 \, \Omega, \, R_3 = 12 \, \Omega \] We need to find the value of \( R_4 \) (which is unknown). Since there is no deflection in the galvanometer, the bridge is balanced. Therefore: \[ \frac{3}{4} = \frac{12}{R_4} \] Solving for \( R_4 \): \[ R_4 = \frac{4 \times 12}{3} = 16 \, \Omega \] Now, using Kirchhoff’s law, we can find the current flowing through the cell. The total resistance in the circuit is the sum of the resistances in series. The total resistance \( R_{\text{total}} \) is: \[ R_{\text{total}} = R_1 + R_2 + R_3 + R_4 = 3 + 4 + 12 + 16 = 35 \, \Omega \] The current flowing through the cell is given by Ohm's law: \[ I = \frac{V}{R_{\text{total}}} = \frac{24}{35} = 0.686 \, \text{A} \] Thus, the current flowing through the cell is 0.686 A.