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Study Abroad Expert | Updated On - Nov 25, 2025
Kirchhoff’s Law is an important chapter in Physics Class 12 for JEE, NEET and Board examinations. The applications of Kirchhoff’s law are crucial in solving circuit-based numerical questions. The German Physicist, Gustav Kirchhoff, introduced Kirchhoff's Laws for the analysis of complex electrical circuits in the year 1845.
- Kirchhoff’s laws are based on the laws of conservation of charge and energy.
- Kirchhoff’s law is a very important topic in Class 12 Chapter 3 Current Electricity.
- The first law of Kirchhoff is also called Kirchhoff’s Current Law or KCL.
- The second law of Kirchhoff is also called Kirchhoff's Voltage Law or KVL.
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What is Kirchhoff's First Law?
According to Kirchhoff’s first law, “The sum of current entering a junction is equal to the sum of current leaving the junction”. Kirchhoff’s first law explains current. Thus, it is also called Kirchhoff’s current law.
What is Kirchhoff's Second Law?
Kirchhoff’s second law states that, “In any closed mesh of an electrical circuit, the algebraic sum of EMFs of the cell and the product of currents and resistance is always equal to zero”. It can be expressed as:
ΣE + ΣIR = 0
Kirchhoff’s second law explains voltage. Thus, it is also called Kirchhoff’s voltage law.
Also Check:
CBSE Class 12 Question Paper (Available)
ISC Class 12 Specimen Paper 2025
Kirchhoff's Law Video Explanation
Terminologies You Should Know Before Solving Questions
- Node: In Physics, a node is a point in a circuit where two or more components are connected. An example of a node can be best understood by the junction of three resistors.
- Branch: A path between two nodes containing a resistor, capacitor, or battery is called a branch in an electrical circuit.
- Loop: A closed path in an electrical circuit where the starting and ending meet at the same point is called a loop.
- Potential Difference: The work done per unit charge between two points is called Voltage or Potential Difference. This is measured in volts.
- EMF (Electromotive Force): The energy supplied per coulomb of charge by a power source, like a battery, is called an Electromotive Force.
Important Questions on Kirchhoff’s Law
Kirchhoff's Laws is an essential chapter in the Class 12 Physics textbook, providing a foundation for understanding complex circuits and solving numerical problems.
- In the JEE Mains, the number of questions asked from current and electricity, especially Kirchhoff's circuit laws, is around 2–3 every year.
- The questions from this section aren't fixed in NEET, but based on previous years' exam experience, we can expect at least one or two questions from the Current and Electricity chapter of Physics.
With a good understanding of Kirchhoff’s laws, you can solve difficult Kirchhoff problems in minutes. Below are some important questions and answers on Kirchhoff’s Law that will help you in strengthening the conceptual clarity and solve numericals while maintaining speed and accuracy.
Short Answer Type Questions (2 Marks)
Ques 1: What sign conventions are used in KVL? (2 marks)
Ans: The understanding of sign conventions is very important to solve numerical problems on Kirchhoff’s Law. The sign convention explains the direction of Voltage and Current in an electrical circuit.
| Movement | Sign Convention |
|---|---|
| Moving from Negative to Positive across a Voltage source | EMF: Positive |
| Moving from Positive to Negative across a Voltage source | EMF: Negative |
Ques 2: If R1 = 2Ω, R2 = 4Ω, and R3 = 6Ω, determine the electric current flows in the circuit below. (2 marks)
Resistor 1 (R1) = 2Ω
Resistor 2 (R2) = 4Ω
Resistor 3 (R3) = 6Ω
Source of emf 1 (E1) = 9V
Source of emf 2 (E2) = 3V
From Kirchhoff’s law, we arrive at the equation,
– I R1 + E1 – I R2 – I R3 – E2 = 0
– 2 I + 9 – 4 I – 6 I – 3 = 0
– 12 I + 6 = 0
– 12 I = – 6
I = -6 / -12
I = 0.5
Thus current flowing through the circuit is 0.5 A.
Ques 3. Determine the electric current that flows in the circuit as shown in the figure below. (2 marks)
Ans. Consider the direction of the current is the same as the direction of clockwise rotation.
- 20 – 5I -5I – 12 – 10I = 0
- 32 – 20I = 0
- 32 = 20I
I = -32 / 20
I = -1.6 A
Since the answer is negative, it means that the direction of the current is actually anti-clockwise.
Ques 4. Determine the electric current that flows in the circuit as shown in the figure below. (2 marks)
Ans. Consider the direction of the current is the same as the direction of clockwise rotation.
– I – 6I + 12 – 2I + 12 = 0
-9I + 24 = 0
-9I = -24
I = 24 / 9
I = 8 / 3 A
Thus, the current flowing through the circuit is 2.66 A.
Long-Answer Type Questions on Kirchhoff’s Law (3 Marks)
Ques 1. An electric circuit consists of four resistors, R1 = 12 Ohms, R2 = 12 Ohms, R3 = 3 Ohms and R4 = 6 Ohms, and are connected with a source of emf E1 = 6 Volt, E2 = 12 Volt. Determine the electric current flows in the circuit as shown in the figure below. (3 marks)
Ans. Resistor 1 (R1) and resistor 2 (R2) are connected in parallel. The equivalent resistor:
1/R12 = 1/R1 + 1/R2 = 1/12 + 1/12 = 2/12
R12 = 12/2 = 6 Ω
In this solution, the direction of the current is the same as the direction of clockwise rotation.
– I R12 – E1 – I R3 – I R4 + E2 = 0
– 6 I – 6 – 3I – 6I + 12 = 0
– 6I – 3I – 6I = 6 -12
– 15I = – 6
I = -6/-15
I = 2/5 A
Thus, the current flowing through the circuit is 0.4 A.
Ques 2: Determine the electric current that flows in the circuit as shown in the figure below. (3 marks)
Ans: Resistor 1 (R1) = 10 Ω
Resistor 2 (R2) = 6 Ω
Resistor 3 (R3) = 5 Ω
Resistor 4 (R4) = 20 Ω
Source of emf 1 (E1) = 8 Volt
Source of emf 2 (E2) = 12 Volt
Resistor 3 (R3) and resistor 4 (R4) are connected in parallel. The equivalent resistor :
1/R34 = 1/R3 + 1/R4 = 1/5 + 1/20 = 4/20 + 1/20 = 5/20
R34 = 20/5 = 4 Ω
In this solution, the direction of the current is the same as the direction of clockwise rotation.
– I R1 – I R2 – E1 – I R34 + E2 = 0
– 10I – 6I – 8 – 4I + 12 = 0
– 10I – 6I – 4I = 8 – 12
– 20I = – 4
I = -4/-20
I = 1/5 A
I = 0.2 A
Thus, the current flowing through the circuit is 0.2 A.
Ques 3. From the given circuit, find the value of I. (3 marks)
Ans. From Kirchhoff’srule,
The arrows pointing toward P are positive, and away from P are negative.
Therefore, 0.2A – 0.4A + 0.6A – 0.5A + 0.7A – I = 0
1.5A – 0.9A – I = 0
0.6A – I = 0
I = 0.6 A
Thus, the current flowing through the circuit is 0.6 A.
Ques 4. The following figure shows a network of conductors. Apply Kirchhoff’s voltage rule to the two closed loops, like ACE and ABC. (3 marks)
Ans. Thus, applying Kirchhoff’s second law to the closed loop EACE,
I1R1 + I2R2 + I3R3 = ξ
For closed loop ABCA,
I4R4 + I5R5 – I2R2= 0
Ques 5. Calculate the current that flows in the 1 Ω resistor in the following circuit. (3 marks)
Ans. Consider the loop EFCBE and apply KVR; we get
1I₂ + 3I₁ + 2I₁ = 9
5I₁ + I₂ = 9 …(1)
Applying KVR to the loop EADFE, we get
3 (I₁ – I₂) – 1I₂ = 6
3I₁ – 4I₂ = 6 …(2)
By solving equations (1) and (2), we get
I1 = 1.83 A and I2 = -0.13 A
The negative sign implies that the current in the 1-ohm resistor flows from F to E.
Ques 6. Find the equivalent resistance using Kirchhoff's laws. (3 marks)
Ans. From the figure, we can write,
6I+3(I+I₂)=E…(1)
3(3−I)+6(3−(I+I₂))=E
9I+3I₂=E
27−9I−6I₂=E
Applying KVL in loop 1,
6I−3(3−I)=0
2I=3−I
3I=3⇒I=1…(2)
Using 2,
9I+3I₂ =27−9I−6I₂
9+3I₂=18−6I₂
9I₂=9
I2=1A
From 1, E=12
R=E/3
R=4Ω
Thus, the equivalent resistance flowing through the circuit is 4Ω.
Ques 7. Prove the conservation of energy by Kirchhoff’s laws. (3 marks)
Ans. According to Kirchhoff’s law, in an electrical circuit, the total power entering the circuit is equal to the total power exiting the circuit.
Consider a circuit consisting of an ideal voltage source with emf ε and a resistance R.
Power in the emf source is Ps =−VI
Power in the resistance is Pr
By Ohm's law, V=IR
⟹Pr =VI
∴Ps +Pr = 0
Ques 8. How to perform Nodal and Mesh Analysis? (3 marks)
Ans. The following are the steps to perform Nodal Analysis based on Kirchhoff’s Current Law.
- Assign an arbitrary value to the voltage at each node in the circuit.
- Apply KCL to each node.
- Calculate branch currents.
- Perform mathematical calculations and obtain the voltage at the node.
The following are the steps to perform Mesh Analysis based on Kirchhoff’s law.
- Every loop is assigned a loop-based current value.
- Kirchhoff’s KVL is implemented for every loop.
- Perform mathematical solvation to obtain the loop current value.
Ques 9. What are the advantages of Kirchhoff’s law in class 12? (3 marks)
Ans. The following are some of the major advantages of Kirchhoff’s law in class 12:
- In calculating unknown voltages and currents in a complete electrical circuit.
- The flow of current and voltage in a circuit can be easily understood with the help of Kirchhoff’s law.
- The complex electric circuit can be studied and simplified by this law.
Very Long Answer Type Questions on Kirchhoff’s Law (5 Marks)
Ques 1. Determine the electric current that flows in the circuit as shown in the figure below. (5 marks)
Ans. Resistor 1 (R1) = 1 Ω
Resistor 2 (R2) = 6 Ω
Resistor 3 (R3) = 6 Ω
Resistor 4 (R4) = 4 Ω
Source of emf 1 (E1) = 12 Volt
Source of emf 2 (E2) = 6 Volt
The electric current that flows in a circuit.
Resistor 1 (R1) and resistor 2 (R2) are connected in parallel. The equivalent resistor :
1/R12 = 1/R1 + 1/R2 = 1/1 + 1/6 = 6/6 + 1/6 = 7/6
R12 = 6/7 Ω
The direction of the current is the same as the direction of clockwise rotation.
E1 – I R12 – E2 – I R4 – I R3 = 0
12 – (6/7)I – 6 – 4I – 6I = 0
12 – 6 – (6/7)I – 4I – 6I = 0
6 – (6/7)I – 10I = 0
6 = (6/7)I + 10I
6 = (6/7)I + (70/7)I
6 = (76/7)I
(6)(7) = 76I
42 = 76I
I = 42/76
I = 0.5 A
Thus, the current flowing through the circuit is 0.5 A.
Ques 2. Why can’t we just use Ohm’s Law instead of Kirchhoff’s Laws? (5 marks)
Ans: Ohm’s Law only tells you the basic V–I–R relationship for a single resistor or a simple part of a circuit. But real exam questions – especially in Class 11, JEE, NEET, and board-level numericals – involve multiple loops, junctions, batteries, and interconnected resistors.
In such cases, current splits, rejoins, and flows through different paths, which Ohm’s Law alone cannot describe.
Kirchhoff’s Laws (KCL & KVL) help you:
- Track how current divides at junctions (KCL)
- Calculate voltage drops and rises across a loop (KVL)
- Solve complex, multi-loop networks that have series + parallel components mixed
- Handle circuits where the total resistance cannot be reduced using simple formulas
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