MP Board · Class 12 · Physics · Current ElectricityState Kirchhoff's rules for electrical networks. Using these rules, obtain the balance condition for a Wheatstone bridge.

Kirchhoff's Rules

\nKirchhoff's rules are essential for analyzing complex electrical circuits where Ohm's single-loop analysis falls short. They are divided into two primary laws:

1. Kirchhoff's First Rule (Current Law or KCL): This rule states that the algebraic sum of currents meeting at any junction in a closed electrical circuit is zero. Mathematically, $\sum I = 0$. This rule is based on the conservation of electric charge. Currents entering the junction are taken as positive, while currents leaving the junction are taken as negative.

2. Kirchhoff's Second Rule (Voltage Law or KVL): This rule states that in any closed loop of a network, the algebraic sum of the products of currents and resistances in various branches of the loop is equal to the algebraic sum of the electromotive forces (EMFs) in that loop. Mathematically, $\sum IR = \sum E$. This rule is based on the law of conservation of energy.

Wheatstone Bridge Principle and Derivation

\nA Wheatstone bridge is an arrangement of four resistances used to measure an unknown resistance accurately. It consists of four resistors $P$, $Q$, $R$, and $S$ connected to form a quadrilateral network (bridge). A source of EMF (battery) is connected across one pair of diagonally opposite corners (say $A$ and $C$), and a sensitive galvanometer is connected across the other pair of opposite corners ($B$ and $D$). \nLet the current from the battery be $I$. At junction $A$, this current splits into $I_1$ (through $P$) and $I_2$ (through $R$). When the bridge is balanced, no current flows through the galvanometer, meaning the current through the galvanometer is zero ($I_g = 0$), and the potentials at points $B$ and $D$ are equal ($V_B = V_D$). \nApplying Kirchhoff's Voltage Law to the closed loop $ABDA$:

• Moving along the path $A \to B \to D \to A$:
• $I_1 P + I_g G - I_2 R = 0$
• Since the bridge is balanced, $I_g = 0$:
• $I_1 P - I_2 R = 0$
• $I_1 P = I_2 R$ \quad (Equation 1) \nApplying Kirchhoff's Voltage Law to the closed loop $BCDB$:
• Moving along the path $B \to C \to D \to B$:
• $(I_1 - I_g)Q - (I_2 + I_g)S - I_g G = 0$
• Since $I_g = 0$:
• $I_1 Q - I_2 S = 0$
• $I_1 Q = I_2 S$ \quad (Equation 2) \nDividing Equation 1 by Equation 2: $$\frac{I_1 P}{I_1 Q} = \frac{I_2 R}{I_2 S}$$ $$\frac{P}{Q} = \frac{R}{S}$$ \nThis is the required balance condition for a Wheatstone bridge. If three resistances are known, the fourth unknown resistance can be easily calculated.