Basic Circuit Analysis Magnetic Circuits Transformers

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

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

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

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) j8 I1 5030 Matrix Form = j8 (6+j4) I2 -80-40

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

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 - (j5 + 6 + j3)I2 = 200 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

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

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 ,

THE IDEAL TRANSFORMER Zin 0 1 0 I1 Vs1 Matrix 0 0 -n 1 V1 0 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

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

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

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

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

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 _ _ _ •

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-148.7 V

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) 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

Basic Laws of Circuits circuits End of Lesson Magnetic Circuits