ELL100: INTRODUCTION TO ELECTRICAL ENG.

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Presentation transcript:

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

Circuit Laws: KCL

Circuit Laws: KVL Branch

Application of Kirchhoff’s Law + + +

Application of Kirchhoff’s Law + + +

Application of Kirchhoff’s Law + + +

Application of Kirchhoff’s Law + + i1+ i2- i3 = 0 (KCL) +

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

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

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

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

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

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

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

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

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

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

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 = 32 (loop 1) 0i1 + 2i2+ 4 i3 = 24 (loop 2)

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 = 32 (loop 1) 0i1 + 2i2+ 4 i3 = 24 (loop 2)

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

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

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

Solution by Matrix algebra method

Solution by Substitution method

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

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

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

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

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

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

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

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

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

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

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