Kirchhoff’s Voltage Law (KVL): Solved Example Problems with Step-by-Step Explanation

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Kirchhoff's Laws – KCL and KVL Solved Examples | BEE | JEE
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Kirchhoff's Voltage Law – Solved Numerical Problems

By Sanket Barde • Basic Electrical Engineering • JEE
Step-by-step KVL numerical problems using mesh-current analysis. Learn how to write loop equations, solve simultaneous equations and determine branch currents.

Solving numerical problems using Kirchhoff's Voltage Law is an essential skill in Basic Electrical Engineering. These solved examples demonstrate how to apply KVL step by step to analyze electrical circuits accurately.

This page is useful for BEE, Diploma, Engineering students and JEE aspirants who want to strengthen their circuit-analysis and problem-solving ability.

Before solving Kirchhoff's Voltage Law numerical problems, make sure you are comfortable with the basic concepts of Kirchhoff's Laws .

1 Kirchhoff's Voltage Law: Solved Example Problems

Q1

Find the current flowing through all resistors.

KVL circuit problem 1
Circuit diagram for Problem 1
✓ Solution
Mesh-1 Equation
\[ 5-2I_1-1(I_1-I_2)-3I_1=0 \]
\[ 5-2I_1-I_1+I_2-3I_1=0 \]
\[ -6I_1+I_2=-5 \]
Mesh-2 Equation
\[ -4I_2-5I_2-1(I_2-I_1)=0 \]
\[ -4I_2-5I_2-I_2+I_1=0 \]
\[ I_1-10I_2=0 \]
Solving the simultaneous equations
\[ I_1=0.8474\,A \] \[ I_2=0.08474\,A \]
✓ Final Answer
\[ I_{2\Omega}=I_{3\Omega}=0.8474\,A \] \[ I_{4\Omega}=I_{5\Omega}=0.08474\,A \] \[ I_{1\Omega}=I_1-I_2=0.7626\,A \]
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Q2

Find the current flowing through the 20 Ω resistor.

KVL circuit problem 2
Circuit diagram for Problem 2
✓ Solution
Mesh-1 Equation
\[ 4-10I_1-5(I_1-I_2)=0 \]
\[ 4-10I_1-5I_1+5I_2=0 \]
\[ -15I_1+5I_2=-4 \]
Mesh-2 Equation
\[ 8-20I_2-10I_2-5(I_2-I_1)=0 \]
\[ 8-30I_2-5I_2+5I_1=0 \]
\[ 5I_1-35I_2=-8 \]
Solving the simultaneous equations
\[ I_1=0.36\,A \] \[ I_2=0.28\,A \]
✓ Final Answer
\[ I_{20\Omega}=0.28\,A \]
Q3

Find the current flowing through the 20 Ω resistor.

KVL circuit problem 3
Circuit diagram for Problem 3
✓ Solution
Mesh-1 Equation
\[ -4-10I_1-5(I_1-I_2)=0 \]
\[ 4-10I_1-5I_1+5I_2=0 \]
\[ -15I_1+5I_2=4 \]
Mesh-2 Equation
\[ 8-5(I_2-I_1)-10I_2-20I_2=0 \]
\[ 8-35I_2+5I_1=0 \]
\[ 5I_1-35I_2=-8 \]
Solving the simultaneous equations
\[ I_1=-0.2\,A \] \[ I_2=0.2\,A \]
✓ Final Answer
\[ I_{20\Omega}=0.2\,A \]
Note: A negative mesh current indicates that the actual current direction is opposite to the initially assumed mesh-current direction.
Q4

Find the current flowing through the 5 Ω resistor.

KVL circuit problem 4
Circuit diagram for Problem 4
✓ Solution
Mesh-1 Equation
\[ 50-2I_1-4(I_1-I_2)=0 \]
\[ 50-2I_1-4I_1+4I_2=0 \]
\[ -6I_1+4I_2=-50 \]
Mesh-2 Equation
\[ -3I_2-5I_2-100-4(I_2-I_1)=0 \]
\[ -3I_2-5I_2-100-4I_2+4I_1=0 \]
\[ 4I_1-12I_2=100 \]
Solving the simultaneous equations
\[ I_1=3.5714\,A \] \[ I_2=-7.1428\,A \]
✓ Final Answer
\[ I_{5\Omega}=-7.1428\,A \]
Interpretation: The negative sign indicates that the actual current direction through the 5 Ω resistor is opposite to the assumed reference direction.
Q5

Find the current flowing through the 2 Ω resistor.

KVL circuit problem 5
Circuit diagram for Problem 5
✓ Solution
Mesh-1 Equation
\[ 16-4I_1-4(I_1-I_2)-12=0 \]
\[ 16-4I_1-4I_1+4I_2-12=0 \]
\[ -8I_1+4I_2=-4 \]
Mesh-2 Equation
\[ -2I_2-1I_2-20+12-4(I_2-I_1)=0 \]
\[ -3I_2-8-4I_2+4I_1=0 \]
\[ 4I_1-7I_2=8 \]
Solving the simultaneous equations
\[ I_1=-0.1\,A \] \[ I_2=-1.2\,A \]
✓ Final Answer
\[ I_{2\Omega}=-1.2\,A \]
Interpretation: The negative sign indicates that the actual current direction is opposite to the assumed direction.

Now that you have studied Kirchhoff's Laws, you may also like to learn how these laws are applied to solve numerical problems using the Superposition Theorem .

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