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Basic Circuit Analysis

Basic Circuit Analysis. Magnetic Circuits Transformers. The Linear Transformer. Illustrating Induced Voltage:. Case 1:. j M. +. +. V 1. I 1. V 2. I 2. •. •. _. _. j L 1. j L 2. V 1 = j L 1 I 1 + jMI 2. V 2 = j L 2 I 2 + jMI 1. The Linear Transformer.

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Basic Circuit Analysis

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  1. Basic Circuit Analysis Magnetic Circuits Transformers

  2. The Linear Transformer Illustrating Induced Voltage: Case 1: jM + + V1 I1 V2 I2 • • _ _ jL1 jL2 V1 = jL1I1 + jMI2 V2 = jL2I2 + jMI1

  3. The Linear Transformer Illustrating Induced Voltage: Case 2: jM _ + • I1 V1 V2 I2 • _ + jL2 jL1 V1 = jL1I1 + jMI2 V2 = jL2I2 + jMI1

  4. The Linear Transformer Illustrating Induced Voltage: Example 1: j8 -j4 6 2 • + + I1 I2 va(t) vb(t) j6 j10 _ _ • va(t) = 50cos(400t + 30) V vb = 80cos(400t – 40) V Va = 50300 V Vb = 80-400 V

  5. EXAMPLE 1: Continued j8 -j4 6  2  • + + I1 I2 j10  j6  80-40 V 5030V _ _ • Solve for I1 and I2 Mesh 1 (2 + j10)I1 + j8I2 = 5030 j8I1 + (j6 – j4 + 6)I2 = - 80-40 Mesh 2 (2+j10) j8 I1 5030 = Matrix Form j8 (6+j4) I2 -80-40

  6. The Linear Transformer Illustrating Induced Voltage: Example 2: 8  -j4  j8  j10  • • j3  + j5  12  200 V I1 _ I2 6  Solve for I1 and I2

  7. Example 2: Continued 8  -j4  j8  j10  • • j3  + j5  12  200 V I1 _ I2 6  (8 + j10 + j5 + 6)I1 - (j5 + 6 + j3)I2 = 200 Mesh 1: Mesh 2 -(6 + j5 + j3)I1 + (6 + j5 + j8 – j4 + 12 + j3)I2 = 0 (14+j15) -(6+j8) I1 200 Matrix = -(6+j8) (18+j12) I2 0

  8. THE IDEAL TRANSFORMER N1 : N2 1 : n ideal If like assumed polarities of the voltages V1 and V2 are placed at the 2dots of the transformer, then V1/V2 = n. If either one of the dots or either one of the voltage polarities are reversed then V2/V1 = -n. If current I1 enters the dot on its side of the transformer and current I2 leaves the dot on its side of the transformer then I1/I2 = n. If either current reverses its direction of entering its respective dot then I1/I2 = - n

  9. THE IDEAL TRANSFORMER 1:n ideal BASIC EQUATIONS: V2 I1 n = = n I2 V1 I2ZL - V2 = -VS2 I1Zin + V1 = VS1 ,

  10. THE IDEAL TRANSFORMER ideal Rearrange previous equations Zin 0 1 0 I1 Vs1 0 ZL 0 -1 I2 -Vs2 = Matrix 0 0 -n 1 V1 0 1 -n 0 0 V2 0

  11. THE IDEAL TRANSFORMER 1 : n ZA ZB VA VB The Basic Transformer Without Markings

  12. THE IDEAL TRANSFORMER • Thevenin Considerations: • • • • or • • this this • • • or • • • this this

  13. THE IDEAL TRANSFORMER 1 : n ZA ZB VA VB ZA ZB n2 _ + + • • • VB VB VA _ • n - n +

  14. THE IDEAL TRANSFORMER 1 : n ZA ZB VA VB ZB n2ZA _ + • • + • nVA nVA VB _ • + _

  15. THE IDEAL TRANSFORMER (4 – j6)  (9 + j18)  1 : 3 ZA ZB _ + • + + 100 V I1 2730 V V1 V2 I2 _ _ + _ • Solve for I1 and I2 Thevenin impedance Thevenin voltage • j2  -j6  1  4  + + + 930 V 100 V I1 V1 _ _ _ •

  16. The Ideal Transformer Thevenin impedance Thevenin voltage • j2  -j6  1  4  + + + 930 V 100 V I1 V1 _ _ _ (5 – j4)I1 = 100 - 930 I1 = 0.78-25 A I1 I2 = - 0.26155 A = 3 10.7131.3 V V1 - 100 + (4 – j6)I1 = 0; V1 = 10 – (4 – j6)(0,78-25 V2 = -3V1 = - 3(10.7131.3) 32.1-148.7 V

  17. The Ideal Transformer (4 – j6)  (9 + j18)  1 : 3 ZA ZB _ + • + + 100 V I1 2730 V V1 V2 I2 _ _ + _ • Non Thevenin Solution: (4 – j6)I1 + V1 = 100 ; (9 + j18)I2 – V2 = 2730 V2 I1 I1 + 3I2 = 0 = - 3 3V1 + V2 = 0 = - 3 V1 I2 (4 – j6) 0 1 0 I1 100 0 (9 + j18) 0 -1 I2 2730 = Matix 0 0 3 1 V1 0 1 3 0 0 V2 0

  18. Basic Laws of Circuits circuits End of Lesson Magnetic Circuits

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