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Basic Circuit Analysis

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Presentation on theme: "Basic Circuit Analysis"— Presentation transcript:

1 Basic Circuit Analysis
Magnetic Circuits Transformers

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

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

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

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

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

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 V2 I1 = = n n V1 I2 I2ZL - V2 = -VS2
BASIC EQUATIONS: V2 I1 = = n n V1 I2 I2ZL - V2 = -VS2 I1Zin + V1 = VS1 ,

10 THE IDEAL TRANSFORMER Zin 0 1 0 I1 Vs1 Matrix 0 0 -n 1 V1 0
Rearrange previous equations Zin I Vs1 ZL I Vs2 = Matrix n V n V

11 THE IDEAL TRANSFORMER The Basic Transformer Without Markings ZA ZB VA
VB The Basic Transformer Without Markings

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

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

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

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

16 The Ideal Transformer Thevenin impedance • Thevenin voltage -j6  1 
4  -j6  1  j2  + + + 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 V1 - 100 + (4 – j6)I1 = 0; V1 = 10 – (4 – j6)(0,78-25 10.7131.3 V V2 = -3V1 = - 3(10.7131.3) 32.1 V

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

18 Basic Laws of Circuits circuits End of Lesson Magnetic Circuits


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