MP Board · Class 12 · Physics · Current ElectricityState Kirchhoff's rules for electrical networks. Using these rules, obtain the balance condition for a Wheatstone bridge. Also, solve a numerical problem where a Wheatstone bridge has arms of resistances $P = 100\,\Omega$, $Q = 10\,\Omega$, $R = 50\,\Omega$, and $S = 5\,\Omega$. Determine the current flowing through the galvanometer of resistance $20\,\Omega$ when a cell of emf $2\,\text{V}$ and negligible internal resistance is connected across the bridge.

Kirchhoff's Rules

1. Kirchhoff's First Rule (Junction Rule / Current Law - KCL): The algebraic sum of currents meeting at any junction in a closed electrical circuit is zero. $$\sum I = 0$$ This rule is based on the conservation of electric charge.

2. Kirchhoff's Second Rule (Loop Rule / Voltage Law - KVL): The algebraic sum of changes in potential around any closed loop involving resistors and cells is zero. $$\sum \Delta V = 0$$ or $\sum E = \sum IR$ This rule is based on the conservation of energy.

Balance Condition for a Wheatstone Bridge

\nA Wheatstone bridge consists of four resistors $P, Q, R,$ and $S$ connected in the form of a quadrilateral. A galvanometer of resistance $G$ is connected between one pair of opposite corners, and a source of emf is connected across the other pair.

• Let the currents through the various arms be applied using Kirchhoff's rules.
• Applying KCL at junction $B$: $I_1 - I_g - I_3 = 0 \implies I_1 = I_3 + I_g$
• Applying KCL at junction $D$: $I_2 + I_g - I_4 = 0 \implies I_4 = I_2 + I_g$
• Applying KVL to the closed loop $ABDA$: $$-I_1 P - I_g G + I_2 R = 0$$
• Applying KVL to the closed loop $BCDB$: $$-I_3 Q + I_4 S + I_g G = 0$$ \nWhen the bridge is balanced, no current flows through the galvanometer, i.e., $I_g = 0$. \nSubstituting $I_g = 0$ in the above equations:
• $I_1 = I_3$ and $I_2 = I_4$
• $I_1 P = I_2 R$
• $I_3 Q = I_4 S \implies I_1 Q = I_2 S$ \nDividing the two equations: $$\frac{I_1 P}{I_1 Q} = \frac{I_2 R}{I_2 S} \implies \frac{P}{Q} = \frac{R}{S}$$\nThis is the required balance condition for a Wheatstone bridge.

Numerical Problem Solution

Given data:

• $P = 100,\Omega$
• $Q = 10,\Omega$
• $R = 50,\Omega$
• $S = 5,\Omega$
• $G = 20,\Omega$
• $\text{EMF of the cell } E = 2,\text{V}$ \nCheck for balance: $\frac{P}{Q} = \frac{100}{10} = 10$ and $\frac{R}{S} = \frac{50}{5} = 10$.\nSince $\frac{P}{Q} = \frac{R}{S}$, the bridge is in a balanced state! Therefore, the potential at point $B$ is equal to the potential at point $D$, meaning no current flows through the galvanometer ($I_g = 0$). \nIf required to calculate total resistance: Equivalent resistance of upper branch ($P+Q$) in parallel with lower branch ($R+S$):
• $R_{\text{upper}} = 100 + 10 = 110,\Omega$
• $R_{\text{lower}} = 50 + 5 = 55,\Omega$
• $R_{\text{equivalent}} = \frac{110 \times 55}{110 + 55} = \frac{6050}{165} = 36.67,\Omega$ \nThus, current through galvanometer is $0,\text{A}$.