Book Page 444-449 Transformers
Review The RMS equivalent current is defined as the dc that will provide the same power in the resistor as the ac does on average P average = I 2 RMS R = 1 2 I 0 2 R= V RMS 2 I RMS = I 0 2 and V RMS = V 0 2 I = I 0 sin ωt and V = V 0 sin ωt R = 1 2 I 0V 0 = I RMS V RMS
Transformer They increase or decrease ac currents and voltages Two coils of wires, primary and secondary To enhance the flux linkage between the primary and secondary coils, transformers are constructed using an iron core. The primary and secondary coils are often concentric The iron core of a transformer is laminated (layered and insulated from other layers) to reduce the eddy current and hysteresis current energy losses. Currents that reduce efficiency = lagging behind, produces heat
Transformer diagram Current flows in primary coil A changing magnetic field is produced Cuts the secondary coil Induces a current Produces a magnetic field in the secondary coil Current falls in primary B collapses Cuts secondary Produces current in secondary in opposite direction V p = N p = I s V s N s I p Step up transformer: N s > N p Step down transformer: N s < N p This is a step-up transformer the emf in the secondary coil is larger than the emf in the primary
Power output ~ 98% due to Eddy current heating in core Transformers efficiency The efficiency of a transformer is energy supplied by secondary coil eff = 100% energy supplied to primary coil Ways to improve efficiency: Laminating core Eddy current produced inside iron core Thin layers of illuminating material separate layers Increases resistance w/o changing magnetic properties current is reduced energy loss reduced from flux and rise in temperature in core
Improving efficiency Core material soft magnetic material such as iron core high flux Easy to magnetize Wires in coil low resistance wires to avoid heating loss Core design such a way that flux cannot leak out as much flux as possible should link both coils so that maximum rate of change of flux occurs
Example The figure below shows a step-down transformer that is used to light a filament globe of resistance 4.0 Ω under operating conditions. Calculate (a) the reading on the voltmeter with S open V p = N p V V s N s = V p N s = 240 50 = 12V s N p 1000 (b) the current in the secondary coil with an effective resistance of 0.2 Ω with S closed R T = 4 + 0.2 = 4.2Ω I = V R = 12 4.2 = 2.86A
continued (c) the power dissipated in the lamp P = I 2 R = 2.86 2 4.0 = 32.7W (d) the power taken from the supply if the primary current is 150 ma P = VI = 240 x 150 x 10 3 = 36W (e) the efficiency of the transformer. P = I 2 R =2.86 2 0.2 = 1.6W eff = energy supplied by secondary coil energy supplied to primary coil 100% = 32.7+1.6 36 100% = 95%
Reasons for power loss Heating effect of current Resistance of metal used Dielectric losses Self-induction Eddy currents opposite direction of induced current causes breaking effect heating effect called iron loss Resistance of wire copper loss
Hysteresis lagging behind Magnetic field strength increases in positive direction, flux density will go down Field strength = zero, but iron core remains magnetized due to flux density Magnetic field is reversed, flux density = 0 In one cycle magnetization lags behind the magnetic field This produces heat iron loss Silicon iron core is used to reduce heat
Flux leakage Physical vibration and noise EM radiation Dielectric loss in insulating material dissipation of energy through movement of charges in an alternating electromagnetic field as polarization switches direction Reasons continued
Transmission of Power Transformers work only if the current is changing; this is one reason why electricity is transmitted as ac. Voltage stepped up to highest possible value To minimize energy losses, the current must kept low
Example An average of 120 kw of power is delivered to a suburb from a power plant 10 km away. The transmission lines have a total resistance of 0.40 Ω. Calculate the power loss if the transmission voltage is (a) 240 V (b) 24 000 V I = P V = 1.2 105 240 = 500A I = P V = 1.2 105 24000 = 5A P lost = I 2 R = 500 2 0.40 P lost = I 2 R = 5 2 0.40 = 100KW =10W Only 20KW reach Only a small amount is consumer lost, most power reaches 100KW are lost consumer
Topic 11: Electromagnetic induction - AHL 11.2 Power generation and transmission Solve problems involving ideal transformers For an ideal transformer there is no power loss between primary and secondary coils. Thus V p I p = V s I s.
Topic 11: Electromagnetic induction - AHL 11.2 Power generation and transmission Solve problems involving ideal transformers V N V s = 2V p. f is not changed in a transformer.
Topic 11: Electromagnetic induction - AHL 11.2 Power generation and transmission Solve problems involving power transmission