# Solved problems on Kirchhoff’s Voltage Law (KVL) / Mesh Analysis

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## Kirchhoff's Voltage Law (KVL) Examples with Solution

1) Find the current flowing through 1 Ω resistance by using Kirchhoff’s voltage law / Mesh analysis. For Mesh 1

$130-0.2I_1-0.2I_1+0.2I_2-110=0$

$-0.4I_1+0.2I_2=-20 \ldots\ldots (1)$

For Mesh 2

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

$0.2I_1-1.4I_2=-110 \ldots\ldots (2)$

$I_1 = 96.1538 A (clockwise)$

$I_2 = 92.3076 A (clockwise)$

$\therefore$ current flowing through 1 Ω resistor is 92.3076 A (clockwise).

2) By using Kirchhoff’s voltage law (KVL) / Mesh analysis find the current flowing through a 5 Ω resistor. For Mesh 1

$2-0.1I_1-10I_1-5I_1+5I_2=0$

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

For Mesh 2

$-0.2I_2-4-5I_2+5I_1-20I_2=0$

$5I_1-25.2I_2=4 \ldots\ldots (2)$

$I_1 = 0.0855 A (clockwise)$

$I_2 = -0.1417 A (clockwise)$

Current flowing through  5 Ω resistor is $I_1-I_2 = 0.0855-(-0.1417) = 0.2272 A (clockwise)$

3) By using Kirchhoff’s voltage law (KVL) / Mesh analysis find the current flowing through all resistances. For Mesh 1

$12.5-I_1-2I_1-3(I_1-I_2)=0$

$-6I_1+3I_2=-12.5 \ldots\ldots (1)$

For Mesh 2

$-2.5I_2-3I_2-3(I_2-I_1)=0$

$3I_1-8.5I_2=0 \ldots\ldots (2)$

$I_1=2.5297 A (clockwise)$

$I_2= 0.8928 A (clockwise)$

$\therefore$ current flowing through 1 Ω resistor is 2.5297 A (clockwise)

$\therefore$ current flowing through 2.5 Ω resistor is 0.8928 A (clockwise)

$\therefore$ current flowing though 2 Ω resistor is 2.5297 A (clockwise)

$\therefore$ current flowing through 3 Ω resistor is 0.8928 A (clockwise)

$\therefore$ current flowing through 3 Ω (middle) resistor is $I_1-I_2 = 2.5297-0.8928=1.6369 A$ (clockwise)

4) By using Kirchhoff’s voltage law (KVL) / Mesh analysis find the current flowing through a 1 Ω resistor. For Mesh 1

$8-2I_1-2(I_1-I_2)-6=0$

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

For Mesh 2

$-I_2-10+6-2(I_2-I_1)=0$

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

$I_1 = -0.25 A (clockwise)$

$I_2 = -1.5 A (clockwise)$

$\therefore$ current flowing through 2 Ω resistor is $I_1 – I_2 = -0.25-(-1.5) = 1.25 A (clockwise)$

5) By using Kirchhoff’s voltage law (KVL) / Mesh analysis find the current flowing through a 4 Ω resistor. For Mesh 1:

$12-2I_1-12(I_1-I_2)-(I_1-I_3)=0$

$-15I_1+12I_2+I_3=-12 \ldots\ldots (1)$

For Mesh 2:

$-10-3(I_2-I_3)-12(I_2-I_1)-2I_2=0$

$12I_1-17I_2+3I_3=10 \ldots\ldots (2)$

For Mesh 3:

$24-(I_3-I_1)-3(I_3-I_2)-4I_3=0$

$I_1+3I_2-8I_3=-24 \ldots\ldots (3)$

$I_1 = 2.7198 A (clockwise)$

$I_2 = 2.05722 A (clockwise)$

$I_3 = 4.1114 A (clockwise)$

$\therefore$ current flowing through 4 Ω resistor is $I_3 = 4.1114 A (clockwise)$

6) By using KVL find Current flowing through 10 Ω resistance. For Mesh 1

$4-5I_1-15(I_1-I_2)=0$

$-20I_1+15I_2=-4 \ldots\ldots (1)$

For Mesh 2

$-10I_2-8(I_2-I_3)-15(I_2-I_1)=0$

$15I_1-32I_2+8I_3=0 \ldots\ldots (2)$

For Mesh 3

$-12I_3-6-8(I_3-I_2)=0$

$8I_2-20I_3=6 \ldots\ldots (3)$

$I_1 =0.2256 A (clockwise)$

$I_2 = 0.0341 A (clockwise)$

$I_3 = -0.2863 A (clockwise)$

$\therefore I_{10Ω} = 0.0341 A (clockwise)$

7) Find current flowing through all resistors.   For Mesh 1:

-2(1-x)-3y+x=0

3x-3y=2 $\ldots\ldots$(1) For Mesh 2:

3y-4(1-x-y)+5(x+y)=0

9x+12y=4  $\ldots\ldots$(2)

x= 0.5714 A

y= -0.0952 A

$I_{1\Omega}$ = x=0.5714 A

$I_{2\Omega}$ = (1-x)=1-0.5714 = 0.4286 A

$I_{3\Omega}$ = 0.0952 A

$I_{4\Omega}$ = (1-x-y) = 1-05714-(-0.0952)=0.5238 A

$I_{5\Omega}$ = x+y = 0.5714 +(-0.0952) = 0.4762 A