# Solved Problems on Superposition Theorem

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1) Find Current flowing through $1 \Omega$ resistor by using superposition theorem.

Consider 130 V Voltage source only.

$-0.2I_1-0.2(I_1-I_2)+130 = 0$

$-0.4I_1 +0.2I_2 = -130 \ldots\ldots (1)$

$-I_1-0.2(I_2-I_1)-0.2I_2=0$

$0.2I_1-1.4I_2=0 \ldots\ldots (2)$

$I_1 = 350 A$

$I_2 = 50 A$

$I’_{1\Omega} = 50 A (\downarrow)$

Consider 110 V only.

$-0.2I_1 -0.2(I_1-I_2)-110 = 0$

$-0.4I_1+0.2I_2 = 110 \ldots\ldots(3)$

$-I_2+110-0.2(I_2-I_1)-0.2I_2=0$

$0.2I_1 – 1.4I_2 = -110 \ldots\ldots (4)$

$I_1 = -253.8467 A$

$I_2 = 42.3076 A$

$I’’_{1\Omega} = 42.3076 A (\downarrow)$

$\therefore I_{1\Omega} = I’_{1\Omega} + I’’_{1\Omega} = 50 + 42.3076 = 92.3076 A (\downarrow)$

2) Find current flowing through 5 $\Omega$.

$2-0.1I_1-10I_1-5(I_1-I_2)=0$

$15.1I_1 + 5I_2=-2 \ldots\ldots (1)$

$-20I_1-0.2I_2-5(I_2-I_1)=0$

$5I_1-25.2I_2=0 \ldots\ldots (2)$

$I_1 = 0.1417 A$

$I_2 = 0.02812 A$

$I’_{5\Omega} = I_1-I_2 = 0.1158 A (\downarrow)$

Consider 4 V only.

$-0.1I_1-10I_1-5(I_1-I_2) = 0$

$-15.1I_1+5I_2 = 0 \ldots\ldots (3)$

$20I_2-0.2I_2-4-5(I_2-I_1)=0$

$5I_1-25.2I_2 = 4 \ldots\ldots (4)$

$I_1 = -0.05685 A$
$I_2 = -0.1698 A$

$I’’_{5 \Omega} = I_2-I_1 = 0.1130 A (\downarrow)$

$I_{5 \Omega} = I’_{5\Omega} + I’’_{5\Omega} = 0.2288 A (\downarrow)$

3) Determine current through 1 $\Omega$ resistor.

Consider a 10 V voltage source only.

$-2(I_1-I_2)-2I_1 = 0$

$-4I_1+2I_2=0 \ldots\ldots (1)$

$-10 -2(I_2-I_1)-I_2=0$

$2I_1-3I_2=10 \ldots\ldots (2)$

$I_1 = -2.5 A$

$I_2 =-5 A$

$I’_{1\Omega} = -5 A (\rightarrow)$

Consider a 6 V voltage source only.

$-2I_1-2(I_1-I_2)-6 = 0$

$-4I_1+2I_2=6 \ldots\ldots (3)$

$-I_2+6-2(I_2-I_1)=0$

$2I_1 – 3I_2 =-6 \ldots\ldots (4)$

$I_1 = -0.75 A$

$I_2 =1.5 A$

$I’’_{1\Omega} = 1.5 A (\rightarrow)$

Consider 8 V only.

$8-2I_1-2(I_1-I_2)=0$

$-4I_1+2I_2=-8 \ldots\ldots (5)$

$-I_2-2(I_2-I_1)=0$

$2I_1-3I_2=0 \ldots\ldots (6)$

$I_1 = 3 A$

$I_2 = 2 A$

$I’”_{1\Omega} = 2 A (\rightarrow)$

$Total \; \; current \; \; I_{1 \Omega}=I’_{1 \Omega}+I’’_{1 \Omega}+I’’’_{1 \Omega}$

$I_{1 \Omega}= 2+1.5+(-5) = -1.5 A (\rightarrow)$

$I_{1 \Omega} = 1.5 A (\leftarrow)$

4) Find current through 20 $\Omega$ resistor.

For Mesh 1:

$10-10I_1+10I_2=0$

$-10I_1+10I_2=-10 \ldots \ldots(1)$

For Mesh 2:

$-10I_2+10I_1-I_2-20I_2=0$

$10I_1-31I_2=0 \ldots\ldots(2)$

$I_1 = 1.4761 A$

$I_2 = 0.4761 A$

$I’_{20\Omega} = 0.4761 A (\downarrow)$

Consider 8 V voltage source only.

$-I-8-20I=0$

$-21I = 8$

$I’’_{20\Omega} = -0.3809 A (\downarrow)$

Consider a 12 V voltage source only.

$-12-8I_1+8I_2=0$

$-8I_1+8I_2=12 \ldots\ldots(3)$

$-I_2-8I_2+8I_1-20I_2=0$

$8I_1-29I_2=0 \ldots\ldots(4)$

$I_1=-2.0714 A$

$I_2=-0.5714 A$

$I’’’_{20\Omega} = -0.5714 (\downarrow)$

$I_{20\Omega} = I’_{20\Omega}+I’’_{20\Omega}+I’’’_{20\Omega}$

$I_{20\Omega} = 0.4761+(-0.3809)+(-0.5714)$

$I_{20\Omega} = -0.4762 A (\downarrow)$

$I_{20\Omega} = 0.4762 A (\uparrow)$