Kirchhoff's voltage law
Kirchhoff's Current Law
Algebraic sum of currents meeting at any junction point in an electric circuit is always zero.
Explanation: Consider a node (O) as shown in Figure Four branches meet at junction or node O.
According to KCL, \[ I_2 + I_4 = I_1 + I_3 \] where \( I_1 \) and \( I_4 \) are incoming currents while \( I_2 \) and \( I_3 \) are outgoing currents.
At any node, \[ \sum \text{Incoming Currents} = \sum \text{Outgoing Currents} \]
Kirchhoff's Voltage Law
In any electrical network, algebraic sum of voltage drops across various elements around any closed loop or mesh is equal to algebraic sum of EMFs in that loop.
Explanation: Consider a circuit as shown in Figure According to KVL,
\[ \sum V = 0 \]
\[ V - IR_1 - IR_2 = 0 \]
\[ V = IR_1 + IR_2 \]
i.e., if we trace any closed path or loop in an electrical network, the algebraic sum of branch voltages is always zero.
Sign Convention:
If you move through the loop in the same direction as the current, the voltage drop is negative.
\[ V = -IR \]
If you move through the loop in the opposite direction of the current, the voltage drop is positive.
\[ V = IR \]
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: \[ I_{2 \Omega} = -1.2~\text{A} \]
Comments
Post a Comment