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Current Electricity - Kirchhoff’s Rules

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Kirchhoff's First Law (Junction Rule): This law is based on the Law of Conservation of Charge. It states that the algebraic sum of currents meeting at any junction in a circuit is zero: ∑I=0\sum I = 0. The sum of currents entering a junction equals the sum of currents leaving it.

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Kirchhoff's Second Law (Loop Rule): This law is based on the Law of Conservation of Energy. It states that the algebraic sum of changes in potential around any closed loop in a circuit is zero: ∑ΔV=0\sum \Delta V = 0.

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Sign Convention for Junction Rule: Currents directed towards the junction are taken as positive (+I+I), and currents directed away from the junction are taken as negative (−I-I).

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Sign Convention for Loop Rule (Potential Drop): When traversing a resistor in the direction of the current, the change in potential is −IR-IR. When traversing against the current, the potential change is +IR+IR.

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Sign Convention for Loop Rule (EMF): When traversing a cell from its negative terminal to its positive terminal, the EMF is taken as positive (+E+E). When traversing from positive to negative, the EMF is taken as negative (−E-E).

📐Formulae

∑I=0\sum I = 0

∑ΔV=0\sum \Delta V = 0

∑E=∑IR\sum E = \sum IR

💡Examples

Problem 1:

Consider a simple closed loop containing a battery of E=12VE = 12V and two resistors R1=2ΩR_1 = 2\Omega and R2=4ΩR_2 = 4\Omega connected in series. Find the current II flowing through the circuit using Kirchhoff's Loop Rule.

Solution:

Let the current in the loop be II. Starting from the negative terminal of the battery and traversing clockwise: 12−I(2)−I(4)=012 - I(2) - I(4) = 0 12−6I=012 - 6I = 0 6I=126I = 12 I=126=2AI = \frac{12}{6} = 2A

Explanation:

According to the Loop Rule, the sum of the EMFs and potential drops must be zero. Since we move from the negative to the positive terminal of the cell, EE is +12V+12V. Since we move in the direction of the current through the resistors, the potential changes are −IR1-I R_1 and −IR2-I R_2.

Problem 2:

At a junction in an electrical circuit, four wires meet. Current I1=5AI_1 = 5A flows towards the junction, I2=2AI_2 = 2A flows away, and I3=4AI_3 = 4A flows towards the junction. Find the magnitude and direction of the current I4I_4 in the fourth wire.

Solution:

Applying Kirchhoff's Junction Rule: ∑Iin=∑Iout\sum I_{in} = \sum I_{out}. Let I4I_4 be directed away from the junction. I1+I3=I2+I4I_1 + I_3 = I_2 + I_4 5+4=2+I45 + 4 = 2 + I_4 9=2+I49 = 2 + I_4 I4=7AI_4 = 7A

Explanation:

By the conservation of charge, the total current entering (5A+4A=9A5A + 4A = 9A) must equal the total current leaving (2A+I42A + I_4). Solving for I4I_4 gives 7A7A flowing away from the junction.

Kirchhoff’s Rules - Revision Notes & Key Formulas | CBSE Class 12 Physics