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Lesson 25: Magnetism and Transformers 1 Learning Objectives • Analyze the relationship between the transformation ratio, voltage ratio, current ratio, and impedance ratio. • Construct a circuit equivalent of a transformer and calculate primary and secondary voltage, current and polarity. • Explain the relationship between the power developed in the primary and secondary of a transformer. 2 Transmission of Power 3 Transformer Overview • A transformer is a magnetically coupled circuit whose operation is governed by Faraday’s Law. 4 Faraday’s Experiment #3: Mutually Induced Voltage • Voltage is induced across Coil 2 when i1 is changing. • When i1 reaches steady state, voltage across Coil 2 returns to zero. − NOTE: The coil to which the source is attached is the primary and the coil attached to the load is the secondary. 5 Transformer Overview • A time-varying current in the primary windings induces a magnetic flux (field) in the iron core. • The flux flows through the core and induces a current the secondary windings. • Thus power flows via the magnetic field without the windings being electrically connected. 6 Winding Direction • The polarity of ac voltages can be changed by changing the direction of the windings. 0º phase shift 180º phase shift 7 Iron-Core Transformers • Two basic types of iron-core transformers are the core type and the shell type. • In both, the core is constructed of laminated sheets of steel to reduce eddy current losses. Core Type 8 Shell Type Iron-Core Transformers • We will consider the ideal transformer which − − − − Neglects coil resistance. Neglects core losses. Assumes all flux is confined to the core. Assumes negligible current required to establish core flux. • Transformer operation is governed by Faraday’s Law. 9 Transformation Ratio • According to Faraday’s Law, voltage (e) is directly related to the number of turns (N) in the primary or secondary windings: dm eN dt • For each winding we can write: dm dm e pri N pri and esec N sec dt dt • Because the flux (m) is the same through both windings, we can write: E pri N pri Esec N sec 10 Transformation Ratio • “The ratio of the primary voltage to secondary voltage is equal to the ratio of primary turns to secondary turns.” E pri Esec N pri N sec • This ratio is called the transformation ratio (or turns ratio) and given by the symbol a. a N pri N sec E pri Esec 11 Step-up and Step-down • Transformers are used to change or “transform” voltage. • Step-Up transformer: − The secondary voltage is higher than the primary voltage. − There are fewer primary windings than secondary windings (a < 1). • Step-Down transformer: − The secondary voltage is lower than the primary voltage. − There are more primary windings than secondary windings (a > 1). 12 Step-up and Step-down Step-Up Step-Down 13 Example Problem 1 Suppose the transformer depicted below has 4000 turns on its primary winding and 1000 turns on its secondary. a. Determine it’s turns ratio (a). Is it step-up or step-down? b. If the primary voltage epri = 480 sin t, what is it’s secondary voltage? a N pri N sec 4000 4 => Step-Down since a > 4 1000 Remember, this is a Step-Down for voltage, but a step-up for current. a E pri Esec Esec E pri a 480V 120V 4 => esec = 120V sin (Ѡt) 14 Current Ratio and Power • Because we are considering an ideal transformer, power in equals power out (Pin = Pout). Sin Sout E pri I pri Esec I sec I pri I sec Esec 1 E pri a • If the voltage is stepped up, then the current is stepped down, and vice versa. 15 Impedance • The impedance of the primary of an ideal transformer is the transformation ratio (a) squared times the impedance of the load (secondary winding) and is derived as below: − NOTE: If the load is capacitive or inductive, the reflected impedance is also capacitive or inductive. Vg VL Np Ns a and Ip Is Ns 1 Np a Recall from Ohm's Law: V V Zp g and ZL L Ip Is Rearranging the equations: Z p a2Z L 16 Example Problem 2 For the figure below Eg = 120 V 0º, the turns ratio is 6:1, and ZLD = 100100j. The transformer is ideal. Find: N a. load voltage b. load current c. generator current d. Active power to the load a) E pri Esec Esec b) I sec N pri N sec 120V 0 6 Esec 1 c) 120V 0 20V 0 6 Esec 20V 0 141mA45 Z sec 100 j100 I pri / g I sec/ LD I pri 1 a I sec 141mA45 23.5mA45 a 6 d) Psec/ LD ( I sec ) 2 RLD (141mA) 2 (100 ) 1.99W Note that Ohm’s law and regular power equations are used here. 17 The Dot Convention • The direction of the windings is not obvious looking at a transformer, therefore we use the dot convention. • Dotted terminals have the same polarity at all instants of time. Used for phase shifting (180). 18 Example Problem 3 For the figure below Ig = 25 30ºmA, the turns ratio is 4:1, and VLD = 600ºV. The transformer is ideal. Find: a. b. c. d. Load current Load impedance Generator voltage Real and Reactive load power a) b) c) d) 19 Example Problem 4 For the figure below i1 = 100 sin (ωt) mA and the transformer is ideal. Determine the secondary currents i2 and i3. Ip Is Ns 1 Np a I1 1 I 2 a * I1 0.25*100mA 25mA I2 a i2 25mA sin(t 180) Why -180⁰ I2 1 I 3 a * I 2 2 * 25mA 50mA I3 a i3 50mA sin(t ) 20 Note the dot convention here, which switches the polarity 180⁰ Power Transformer Ratings • Just like ac motors and generators, power transformers are rated in terms of voltage and apparent power. − For example, a transformer is rated at 2400/120 volt, 48 kVA has: S recall: I E • On the primary winding, the current rating is 48,000 VA / 2400 V = 20 A. • On the secondary winding, the current rating is 48,000 VA / 120 V = 400 A. 21 Example Problem 5 A 7.2 kV, a=0.2 transformer has a secondary winding rated current of 3 A. What is its kVA rating? I pri I sec 1 I 3A I pri sec 15 A a a 0.2 Sin E pri * I pri Sin / g I pri / g * E pri / g 15 A * 7.2kV 108kVA 22 Transmission of Power 23 Transformers Mitsubishi 500 MVA Single-Phase Auto-Transformers 24 QUESTIONS? 25