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Consider a simple power system consisting of an ideal voltage source, an ideal step-up transformer, a transmission line, an ideal step-down transformer, and a load. The voltage of the source is
The impedance of the transmission line is
- Assume that the transformers are not present in the circuit. What is the load voltage and efficiency of the system?
- Assume that transformer I is a 1:5 step-up transformer, and transformer 2 is a 5:1 step-down transformer. What is the load voltage and efficiency of the system?
- What transformer turns ratio would be required to reduce the transmission line losses to 1% of the total power produced by the generator?
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Chapter 2 Solutions
Electric machinery fundamentals
- A generator (480 V, 10 kVA) is connected to an ideal step-up transformer (1:20), and a transmission line (R=20 Ohm and X-60 Ohm), the base values for this system are chosen to be 480 V and 10 kVA at the generator. The line base impedance is: Select one: O a. None of the above O b. 9216 Om O c. 1.25 Ohm O d. 13 Ohm O e. 520 Ohmarrow_forwardAn ideal transformer is connected to a 110V(rms), 60HZ source. Line impedance on the primary is 4+5i2 and on the secondary side is 1+2i2. A lmF capacitor is connected as load on the secondary side. Transformer has 100 turns on the primary side and 50 turns on the secondary side. Calculate: (i) Primary and secondary current (ii) Active and reactive power supplied by the source. (iii)Power factor of the source.arrow_forwardConsider an ac network with an ideal transformer (with turns ratio of n:1) shown below where ZL=(1+j0) is the load impedance and Zs=(7−j24). Ideal transformer turns ratio n is chosen so that the average power delivered to the load is maximized. Under this scenario, the average power delivered to the load is given byarrow_forward
- generator (b) What is the average power dissipated in R? P=[ W Help Enter 500 v generator voltage -500 v AC Generator 0 Ideal Transformer R = 300 Rg8 time (a) The sinusoidal voltage output from the AC generator oscillates between 500 V and -500 V. The circuit for an ideal transformer is constructed with a generator, transformer and a 30 W resistor. The generator makes 6 loops with the transformer and the resistor makes 11 loops with the transformer (the figure does not show these loops accurately). What is the maximum voltage across the resistor R? V= V Help Enterarrow_forwardConsider the circuit below: 34j 0 1200:2400 -10j 2 ) 1200Z 30° V 50 Q 100 Q -3j Q The transformer rating is 1200:2400 V @ 120 kVA. What is the apparent power provided by the source? What does this mean for the operation of the transformer? The magnitude of the voltage source is given in VRMS.arrow_forwardGiven Zo-j407 2, R₁143, R6317 52 and a 0.04, what is the magnitude of the source voltage Vs (in V rms) in the primary of the ideal transformer? Vs R₁ ww Answer: Zc Zc R₁ Use up to 2 digits after the decimal point if necessary in your answer below VL-28020 V TRarrow_forward
- In the circuit shown below, the transformer is driven by a source with an open circuit voltage Vs of 120 V (rms) and an internal impedance Zs of 6 22. The frequency of the source is 200 rad/sec. The transformer drives a load consisting of a 60 n resistor. The transformer parameters are: R₁ = 12 Q2, R2 = 24 2, L₁ = 30 mH, L2 = 120 mH and k = 0.9. L1 Determine the currents /1 and /2 and the power delivered to the load. Vs Zs I₁ a b R₁ M L₁ M L₂ R₂ m d 1₂ N ZLarrow_forwardA 40-kVA, 8000/230-V distribution transformer has an impedance referred to the primary of 18 + j80 Q. The components of the excitation branch referred to the primary side are Rc = 100 k2 and Xc - 20 kΩ If the primary voltage is 7920 V and the load impedance is ZL = 1.5 + j*0.7 Q what is the secondary voltage, in volts?arrow_forwardA transformer that can be considered ideal has 200 turns on its primary winding and 500 turns on its secondary winding. The primary is connected to a 220-V sinusoidal supply and the secondary supplies 10 kVA to a load. a. Determine the load voltage, secondary current, and primary current. b. Find the magnitude of the load impedance as seen from the supply.arrow_forward
- 1. A single-phase transformer has Z 1 = 1.4+j5.2Ω and Z 2 = 0.0117+j0.0465Ω. Theinput voltage is 6600V and the turn ratio is 10.6:1. The secondary feeds a loadwhich draws 300A at 0.8 power factor lagging. Find the secondary terminalvoltage and the kW output. Neglect no-load current.arrow_forwardQ2: A 1OKVA distribution transformer has a full.load e fficiency at 0-85 p.f of 95/ the ca ppes and van losses then being equal. Caleulote its all- day efficiency through out the 24 hours if it is loaded as follows Neload Quarter lod lo hours 5 hours 子 howe 2 heurs load Pf .f 0.9 half load full load Assumsarrow_forward10-KVA, 2400/240 V, single-phase transformer has a secondary equivalent impedance of (0.06 + 10.30) ohms. Find the primary voltage required to produce 240 V at the secondary terminals at full load, 0.8 pf leading. A. 2356 VB. 2348 V C. 2372 V D. 2364 Varrow_forward
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