Kirchhoff’s Voltage Laws Solved Examples
Kirchhoff’s Voltage Law (KVL) – Solved Numerical Problems
Solving numerical problems using Kirchhoff’s Voltage Law is an essential skill in Basic Electrical Engineering. These solved examples will help you understand 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 problem-solving ability.
Before solving Kirchhoff’s Voltage Law numerical problems, make sure you are comfortable with the following basic concepts: Kirchhoff’s laws
Kirchhoff's Voltage law: Solved Example Problems
Q.1 Find current flowing through all resistors
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 above simultaneous equations:
\[ I_1 = 0.8474~\text{A} \]
\[ I_2 = 0.08474~\text{A} \]
Answer: \[ I_{2 \Omega} = I_{3 \Omega} = 0.8474~\text{A} \] \[ I_{4 \Omega} = I_{5 \Omega} = 0.08474~\text{A} \] \[ I_{1 \Omega} = I_1 - I_2 = 0.7626~\text{A} \]
Q.2 Find current flowing through 20 Ohm resistors
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 above simultaneous equations:
\[ I_1 = 0.36~\text{A} \]
\[ I_2 = 0.28~\text{A} \]
Answer: \[ I_{20 \Omega} = 0.28~\text{A} \]
Q.3 Find current flowing through 20 Ohm resistors
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 above simultaneous equations:
\[ I_1 =-0.2~\text{A} \]
\[ I_2 = 0.2~\text{A} \]
Answer: \[ I_{20 \Omega} = 0.2~\text{A} \]
Q.4 Find current flowing through 5 Ohm resistors
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 above simultaneous equations:
\[ I_1 = 3.5714~\text{A} \]
\[ I_2 = -7.1428~\text{A} \]
Answer: \[ I_{5 \Omega} = -7.1428~\text{A} \]
Q.5 Find current flowing through 2 Ohm resistors
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 above simultaneous equations:
\[ I_1 = -0.1~\text{A} \]
\[ I_2 = -1.2~\text{A} \]
Answer: \[ I_{2 \Omega} = -1.2~\text{A} \]
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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