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ELL100: INTRODUCTION TO ELECTRICAL ENG.

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Presentation on theme: "ELL100: INTRODUCTION TO ELECTRICAL ENG."— Presentation transcript:

1 ELL100: INTRODUCTION TO ELECTRICAL ENG.
Lecture 2 Course Instructors: J.-B. Seo, S. Srirangarajan, S.-D. Roy, and S. Janardhanan Department of Electrical Engineering, IITD

2 Circuit Laws: KCL

3 Circuit Laws: KVL Branch

4 Application of Kirchhoff’s Law
+ + +

5 Application of Kirchhoff’s Law
+ + +

6 Application of Kirchhoff’s Law
+ + +

7 Application of Kirchhoff’s Law
+ + i1+ i2- i3 = (KCL) +

8 Application of Kirchhoff’s Law
+ + i1+ i2- i3 = (KCL) + 32 - 3i1- 4 i3 = (loop 1) 24 - 2i2- 4 i3 = (loop 2) Rearranging, we get i1+ i2- i3 = 0 3i1+ 0 i2 + 4 i3 = (loop 1) 0i1 + 2i2+ 4 i3 = (loop 2)

9 Application of Kirchhoff’s Law
+ + i1+ i2- i3 = (KCL) + 32 - 3i1- 4 i3 = (loop 1) 24 - 2i2- 4 i3 = (loop 2) Rearranging, we get i1+ i2- i3 = 0 3i1+ 0 i2 + 4 i3 = (loop 1) 0i1 + 2i2+ 4 i3 = (loop 2)

10 Application of Kirchhoff’s Law
+ + i1+ i2- i3 = (KCL) + 32 - 3i1- 4 i3 = (loop 1) 24 - 2i2- 4 i3 = (loop 2) Rearranging, we get i1+ i2- i3 = 0 3i1+ 0 i2 + 4 i3 = (loop 1) 0i1 + 2i2+ 4 i3 = (loop 2)

11 Application of Kirchhoff’s Law
+ + i1+ i2- i3 = (KCL) + 32 - 3i1- 4 i3 = (loop 1) 24 - 2i2- 4 i3 = (loop 2) Rearranging, we get i1+ i2- i3 = 0 3i1+ 0 i2 + 4 i3 = (loop 1) 0i1 + 2i2+ 4 i3 = (loop 2)

12 Application of Kirchhoff’s Law
+ + i1+ i2- i3 = (KCL) + 32 - 3i1- 4 i3 = (loop 1) 24 - 2i2- 4 i3 = (loop 2) Rearranging, we get i1+ i2- i3 = 0 3i1+ 0 i2 + 4 i3 = (loop 1) 0i1 + 2i2+ 4 i3 = (loop 2)

13 Application of Kirchhoff’s Law
+ + i1+ i2- i3 = (KCL) + 32 - 3i1- 4 i3 = (loop 1) 24 - 2i2- 4 i3 = (loop 2) Rearranging, we get Coefficient matrix i1+ i2- i3 = 0 3i1+ 0 i2 + 4 i3 = (loop 1) 0i1 + 2i2+ 4 i3 = (loop 2)

14 Determinant of a matrix
(2 x 2) matrix (3 x 3) matrix

15 Cramer's method – linear algebra
Use determinants to solve the linear equations

16 Application of Kirchhoff’s Law
+ + i1+ i2- i3 = (KCL) + 32 - 3i1- 4 i3 = (loop 1) 24 - 2i2- 4 i3 = (loop 2) Rearranging, we get Coefficient matrix i1+ i2- i3 = 0 3i1+ 0 i2 + 4 i3 = (loop 1) 0i1 + 2i2+ 4 i3 = (loop 2)

17 Solution by Determinant method
Replace the first column with the right side of previous equation (Cramer’s rule)

18 Solution by Determinant method
Replace the first column with the right side of previous equation (Cramer’s rule) i1+ i2- i3 = 0 3i1+ 0 i2 + 4 i3 = (loop 1) 0i1 + 2i2+ 4 i3 = (loop 2)

19 Solution by Determinant method
Replace the first column with the right side of previous equation (Cramer’s rule) i1+ i2- i3 = 0 3i1+ 0 i2 + 4 i3 = (loop 1) 0i1 + 2i2+ 4 i3 = (loop 2)

20 Solution by Determinant method
Replace the first column with the right side of previous equation (Cramer’s rule)

21 Solution by Determinant method
Replace the first column with the right side of previous equation (Cramer’s rule)

22 Solution by Matrix algebra method
+ + i1+ i2- i3 = (KCL) + 32 - 3i1- 4 i3 = (loop 1) 24 - 2i2- 4 i3 = (loop 2) Rearranging, we get i1+ i2- i3 = 0 3i1+ 0 i2 + 4 i3 = (loop 1) 0i1 + 2i2+ 4 i3 = (loop 2)

23 Solution by Matrix algebra method

24 Solution by Substitution method

25 Solution by Loop current method
Just assume single currents In in each loop and apply the KVL to individual loops a b c d

26 Solution by Loop current method
Just assume single currents In in each loop and apply the KVL to individual loops a b c d

27 Solution by Loop current method
Just assume single currents In in each loop and apply the KVL to individual loops a b c d

28 Solution by Loop current method
Just assume single currents In in each loop and apply the KVL to individual loops 32 - 3i1 - 4 (i1 + i2 ) = 0 a b c d

29 Solution by Loop current method
Just assume single currents In in each loop and apply the KVL to individual loops 32 - 3i1 - 4 (i1 + i2 ) = 0 a b c d

30 Solution by Loop current method
Just assume single currents In in each loop and apply the KVL to individual loops 32 - 3i1 - 4 (i1 + i2 ) = 0 24 - 2i2 - 4 (i1 + i2 ) = 0 a b c d

31 Solution by Loop current method
Just assume single currents In in each loop and apply the KVL to individual loops 32 - 3i1 - 4 (i1 + i2 ) = 0 24 - 2i2 - 4 (i1 + i2 ) = 0 a b c d

32 Solution by Node Voltage Method
This method is helpful in reducing the number of variables. a b c d

33 Solution by Node Voltage Method
This method is helpful in reducing the number of variables. a b c d

34 Solution by Node Voltage Method
This method is helpful in reducing the number of variables. a b c d

35 Solution by Node Voltage Method
This method is helpful in reducing the number of variables. a b c d


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