Kirchhoff's Voltage law examples with solution

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.

KVL solved problems

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.

kvl solved problems

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.

Mesh Analysis solved examples

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.

KVL solved examples

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.

kvl solved problems

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.

KVL Solved problems

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.

kvl solved problem
kvl solved problem
kvl solved problem

For Mesh 1:

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

3x-3y=2 \ldots\ldots (1)

kvl solved problem

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

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