Kirchhoff's voltage law
Superposition Theorem
📌Statement: In any linear, bilateral network containing atleast two energy sources, the cur rent flowing through a particular branch is the algebraic sum of the currents flow ing through that branch when each source is considered separately and remaining sources are replaced by their respective internal resistances.
Explanation:
Consider the circuit shown in the above figure. We need to estimate the current \( I \) through \( R_L \) using the Superposition Theorem.
Consider voltage source \( V_1 \) acting alone. Remove the other voltage source i.e. replace it by its internal resistance. As the internal resistance of an ideal voltage source is zero, it is replaced by a short circuit. The current through \( R_L \) is \( I' \). (Refer below circuit)
Now consider the second voltage source \( V_2 \) acting alone. Replace the first voltage source with a short circuit. The current through \( R_L \) is \( I'' \). (Refer below circuit)
\( I' \) and \( I'' \) can be calculated by using Kirchhoff’s Voltage Law.
Therefore, the current flowing through \( R_L \) when both the voltage sources are acting is:
\[ I = I' + I'' \]
Superposition Theorem: Solved Example Problems
Q.1 Find the current flowing through the 20 Ω resistance.
Consider Only 4 V source active
🔁 Mesh–1 Equation:
\[ 4 - 10I_1 - 5(I_1 - I_2) = 0 \]
\[ -15I_1 + 5I_2 = -4 \]
🔁 Mesh–2 Equation:
\[ -20I_2 - 10I_2 - 5(I_2 - I_1) = 0 \]
\[ 5I_1 - 35I_2 = 0 \]
\[ I_1 = 0.28~\text{A} \] \[ I_2 = 0.04~\text{A} \] \[ I'_{20\Omega} = 0.04~\text{A} \;(\rightarrow) \]
Consider Only 8 V source active
🔁 Mesh–1 Equation:
\[ -10I_1 - 5(I_1 - I_2) = 0 \]
\[ -15I_1 + 5I_2 = 0 \]
🔁 Mesh–2 Equation:
\[ 8 - 20I_2 - 10I_2 - 5(I_2 - I_1) = 0 \]
\[ 5I_1 - 35I_2 = -8 \]
\[ I_1 = 0.08~\text{A} \] \[ I_2 = 0.24~\text{A} \] \[ I''_{20\Omega} = 0.24~\text{A} \;(\rightarrow) \]
\[ I_{20\Omega} =I'_{20 \Omega} + I''_{20 \Omega} \] \[ I_{20\Omega} = 0.04 + 0.24 \] \[ I_{20\Omega} = \boxed{0.28~\text{A}} \;(\rightarrow) \]
Q.2 Find the current flowing through the 10 Ω resistance.
Consider Only 4 V source active
🔁 Mesh–1 Equation:
\[ -4 - 10I_1 - 5(I_1 - I_2) = 0 \]
\[ -15I_1 + 5I_2 = 4 \]
🔁 Mesh–2 Equation:
\[ -20I_2 - 10I_2 - 5(I_2 - I_1) = 0 \]
\[ 5I_1 - 35I_2 = 0 \]
\[ I_1 = -0.28~\text{A} \] \[ I_2 = -0.04~\text{A} \] \[ I'_{20\Omega} = -0.04~\text{A} \;(\rightarrow) \]
Consider Only 8 V source active
🔁 Mesh–1 Equation:
\[ -10I_1 - 5(I_1 - I_2) = 0 \]
\[ -15I_1 + 5I_2 = 0 \]
🔁 Mesh–2 Equation:
\[ 8 - 20I_2 - 10I_2 - 5(I_2 - I_1) = 0 \]
\[ 5I_1 - 35I_2 = -8 \]
\[ I_1 = 0.08~\text{A} \] \[ I_2 = 0.24~\text{A} \] \[ I''_{20\Omega} = 0.24~\text{A} \;(\rightarrow) \]
\[ I_{20\Omega} =I'_{20 \Omega} + I''_{20 \Omega} \] \[ I_{20\Omega} = -0.04 + 0.24 \] \[ I_{20\Omega} = \boxed{0.2~\text{A}} \;(\rightarrow) \]
Q.3 Find the current flowing through the 5 Ω resistance.
Consider Only 50 V source active
🔁 Mesh–1 Equation:
\[ 50 - 2I_1 - 4(I_1 - I_2) = 0 \]
\[ -6I_1 + 4I_2 = -50 \]
🔁 Mesh–2 Equation:
\[ -3I_2 -5I_2 - 4(I_2-I_1) = 0 \]
\[ 4I_1 -12I_2 = 0 \]
\[ I_1 = 10.7142~\text{A} \] \[ I_2 = 3.5714~\text{A} \] \[ I'_{5\Omega} = 3.5714~\text{A} \;(\downarrow) \]
Consider Only 100 V source active
🔁 Mesh–1 Equation:
\[ -2I_1 - 4(I_1 - I_2) = 0 \] \[ -6I_1 + 4I_2 = 0 \]
🔁 Mesh–2 Equation:
\[ -3I_2 -5I_2 - 4(I_2-I_1) -100= 0 \] \[ 4I_1 -12I_2 = 100 \]
\[ I_1 = -7.1428~\text{A}\] \[ I_2 = -10.7142~\text{A}\]
\[ I''_{5 \Omega} = -10.7142~\text{A} (\downarrow) \]
\[ I_{5\Omega} =I'_{5 \Omega} + I''_{5 \Omega} \] \[ I_{20\Omega} = 3.5714 + (-10.7142) \] \[ I_{20\Omega} = \boxed{-7.1428~\text{A}} \;(\downarrow) \] \[ I_{20\Omega} = \boxed{7.1428~\text{A}} \;(\uparrow) \]
Q.4 Find the current flowing through the 1 Ω resistance.
Consider Only 16 V source active
🔁 Mesh–1 Equation:
\[ 16 -4I_1 -4(I_1 - I_2) = 0 \] \[ -8I_1 + 4I_2 = -16 \]
🔁 Mesh–2 Equation:
\[ -2I_2 -1I_2 -4(I_2 - I_1) = 0 \] \[ 4I_1-7I_2=0 \]
\[ I_1 = 2.8~\text{A}\] \[ I_2 = 1.6~\text{A}\]
\[ I'_{2 \Omega} = 1.6~\text{A} (\rightarrow) \]
Consider Only 12 V source active
🔁 Mesh–1 Equation:
\[ -4I_1 -4(I_1 - I_2)-12 = 0 \] \[ -8I_1 + 4I_2 = 12 \]
🔁 Mesh–2 Equation:
\[ -2I_2 -1I_2 +12 -4(I_2 - I_1) = 0 \] \[ 4I_1-7I_2=-12 \]
\[ I_1 = -0.9~\text{A}\] \[ I_2 = 1.2~\text{A}\]
\[ I''_{2 \Omega} = 1.2~\text{A} (\rightarrow) \]
Consider Only 20 V source active
🔁 Mesh–1 Equation:
\[ -4I_1 -4(I_1 - I_2) = 0 \] \[ -8I_1 + 4I_2 = 0 \]
🔁 Mesh–2 Equation:
\[ -2I_2 -1I_2 -20 -4(I_2 - I_1) = 0 \] \[ 4I_1-7I_2=20 \]
.\[ I_1 = -2~\text{A}\] \[ I_2 = -4~\text{A}\]
\[ I'''_{2 \Omega} = -4~\text{A} (\rightarrow) \]
\[ I_{2\Omega} = I'_{2\Omega}+I''_{2\Omega}+I'''_{2\Omega} \] \[ I_{2\Omega} =1.6 +1.2+(-4) \] \[ I_{2\Omega} =-1.2 (\rightarrow) \] \[ I_{2\Omega} =1.2 (\leftarrow) \]
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