Question

State and explain Kirchhoff's laws. Applying these laws, obtain the balanced condition of Wheatstone bridge.

Hint:

For Kirchhoff's laws, consistency in your sign convention is key. Choose a direction for your loop traversal and stick to it. For the Wheatstone bridge, the balanced condition \(I_g=0\) is the crucial step that simplifies the loop equations.

Answer 1

Part 1: Kirchhoff's Laws
Step 1: Kirchhoff's First Law (The Junction Rule or Current Law - KCL):
Statement: The algebraic sum of the electric currents meeting at any junction in an electrical circuit is zero.
\[ \sum I = 0 \] Explanation: This law is based on the law of conservation of charge. A junction is a point in a circuit where charge cannot accumulate. Therefore, the total current flowing into the junction must be equal to the total current flowing out of the junction. By convention, currents entering a junction are taken as positive, and currents leaving are taken as negative.
For example, at a junction where currents \(I_1\) and \(I_2\) enter and \(I_3\) and \(I_4\) leave, we have: \(I_1 + I_2 - I_3 - I_4 = 0\).
Step 2: Kirchhoff's Second Law (The Loop Rule or Voltage Law - KVL):
Statement: In any closed loop or mesh of an electrical circuit, the algebraic sum of the changes in potential (products of current and resistance) is equal to the algebraic sum of the electromotive forces (e.m.f.s) in that loop.
\[ \sum IR = \sum E \quad \text{or} \quad \sum \Delta V = 0 \] Explanation: This law is based on the law of conservation of energy. If we start at any point in a closed loop and travel around it, the electric potential must return to its initial value. This means the total potential gained (from batteries) must equal the total potential dropped (across resistors).
Sign Convention:
- A rise in potential (moving from - to + terminal of a cell) is taken as positive e.m.f.
- A fall in potential (moving from + to - terminal) is taken as negative e.m.f.
- A potential drop across a resistor in the direction of current is taken as negative (\(-IR\)).
- A potential gain across a resistor against the direction of current is taken as positive (\(+IR\)).
Part 2: Balanced Condition of Wheatstone Bridge
Step 1: Circuit Diagram:
A Wheatstone bridge consists of four resistors P, Q, R, and S arranged in a quadrilateral ABCD. A cell of e.m.f. E is connected between points A and C, and a galvanometer of resistance G is connected between B and D.

Step 2: Applying Kirchhoff's Laws:
Let the current from the cell be \(I\). At junction A, it splits into \(I_1\) (through P) and \(I_2\) (through R). At junction B, \(I_1\) splits into \(I_g\) (through G) and \(I_1 - I_g\) (through Q). At junction D, currents \(I_2\) and \(I_g\) combine to flow through S.
Applying Kirchhoff's Loop Rule to the closed loop ABDA:
\[ -I_1 P - I_g G + I_2 R = 0 \quad \cdots(1) \] Applying Kirchhoff's Loop Rule to the closed loop BCDB:
\[ -(I_1 - I_g)Q + (I_2 + I_g)S + I_g G = 0 \quad \cdots(2) \] Step 3: Deriving the Balanced Condition:
The bridge is said to be balanced when there is no current flowing through the galvanometer. This means the potential at point B is equal to the potential at point D (\(V_B = V_D\)).
The condition for balance is \(I_g = 0\).
Substituting \(I_g = 0\) into equation (1):
\[ -I_1 P + I_2 R = 0 \implies I_1 P = I_2 R \quad \cdots(3) \] Substituting \(I_g = 0\) into equation (2):
\[ -I_1 Q + I_2 S = 0 \implies I_1 Q = I_2 S \quad \cdots(4) \] Now, divide equation (3) by equation (4):
\[ \frac{I_1 P}{I_1 Q} = \frac{I_2 R}{I_2 S} \] \[ \frac{P}{Q} = \frac{R}{S} \] This is the required balanced condition for a Wheatstone bridge.