Concept: • A Wheatstone bridge consists of four resistance arms $P, Q, R, S$ forming a closed loop $ABCD$, with a voltage source $V$ connected across $AC$ and a galvanometer $G$ across $BD$.
• Balance condition corresponds to zero current through the galvanometer ($I_g = 0$), implying $V_B = V_D$. Step 1: Circuit Diagram and Kirchhoff's Current Law
Let current $I$ from battery split at node $A$ into $I_1$ through arm $AB$ (resistance $P$) and $I_2$ through arm $AD$ (resistance $R$).
At node $B$, current $I_g$ flows through galvanometer arm $BD$ (resistance $G$). Current through arm $BC$ (resistance $Q$) is $I_1 - I_g$.
At node $D$, current through arm $DC$ (resistance $S$) is $I_2 + I_g$. Step 2: Apply Kirchhoff's Voltage Law (KVL)
Apply KVL to closed loop $ABDA$:
\[ -I_1 P - I_g G + I_2 R = 0 \implies I_1 P + I_g G = I_2 R \quad \text{--- (Equation 1)} \]
Apply KVL to closed loop $BCDB$:
\[ -(I_1 - I_g) Q + (I_2 + I_g) S + I_g G = 0 \quad \text{--- (Equation 2)} \] Step 3: Apply Balance Condition ($I_g = 0$)
For balanced bridge, potential at node $B$ equals potential at node $D$ ($V_B = V_D$), so zero current passes through galvanometer ($I_g = 0$).
Substitute $I_g = 0$ into Equation 1:
\[ I_1 P = I_2 R \quad \text{--- (Equation 3)} \]
Substitute $I_g = 0$ into Equation 2:
\[ I_1 Q = I_2 S \quad \text{--- (Equation 4)} \] Step 4: Divide Equation 3 by Equation 4
\[ \frac{I_1 P}{I_1 Q} = \frac{I_2 R}{I_2 S} \]
Canceling non-zero current factors $I_1$ and $I_2$:
\[ \frac{P}{Q} = \frac{R}{S} \] Step 5: Conclusion
The balance condition for a Wheatstone bridge is $\frac{P}{Q} = \frac{R}{S}$.
