Question

Derive the balance condition of a Wheatstone Bridge.

Hint:

Wheatstone bridge balance condition: \[ \frac{P}{Q} = \frac{R}{S} \] At balance → galvanometer shows zero deflection.

Answer 1

Concept:
• A Wheatstone bridge has four resistances \(P, Q, R, S\) forming the arms \(AB, BC, AD, DC\); a battery is joined across \(A\) and \(C\) and a galvanometer \(G\) across \(B\) and \(D\).

• The bridge is balanced when the galvanometer current \(I_g = 0\), that is, when the potentials of \(B\) and \(D\) are equal.

Step 1: Currents in the two branches
Let the battery current split at \(A\) into \(I_1\) through \(P\) (arm \(AB\)) and \(I_2\) through \(R\) (arm \(AD\)).
When \(I_g = 0\), no current enters or leaves the galvanometer branch, so the same \(I_1\) flows through \(Q\) (arm \(BC\)) and the same \(I_2\) flows through \(S\) (arm \(DC\)).

Step 2: Equal potentials at \(B\) and \(D\)
Since \(V_B = V_D\), the potential drops from \(A\) to \(B\) and from \(A\) to \(D\) are equal:
\[ I_1 P = I_2 R \quad \text{--- (1)} \]
Likewise the drops from \(B\) to \(C\) and from \(D\) to \(C\) are equal:
\[ I_1 Q = I_2 S \quad \text{--- (2)} \]

Step 3: Divide equation (1) by equation (2)
\[ \frac{I_1 P}{I_1 Q} = \frac{I_2 R}{I_2 S} \]
Cancelling \(I_1\) and \(I_2\):
\[ \frac{P}{Q} = \frac{R}{S} \]

Final Answer:
The Wheatstone bridge is balanced when \(\dfrac{P}{Q} = \dfrac{R}{S}\), that is, \(PS = QR\).