
ANALYSIS+DESIGN OF LINEAR CIRCUITS(LL)
8th Edition
ISBN: 9781119235385
Author: Thomas
Publisher: WILEY
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4. A 120/240 volt 10kVA two-winding transformer is to be reconnected as a step-up
120/360 volt auto-transformer. What is the kVA rating of the new auto-
transformer connection?
(10 pts)
Solve this problem and show all of the work
3. The total power consumed by a balanced three-phase 480 volts rms line-to-line
system is 13.3 kW and 9.97 kVARS. Determine the current (both magnitude and
phase) supplied to the load.
(10 pts)
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- 2. Determine the rms current (magnitude and phase) supplied by the 120 volt rms source in the circuit shown below. Assume the transformer is ideal. (20 pts) 120 Loo Volts rms. j452 1:3 272arrow_forward6. Find the transfer function H(jw) = Vo(jw)/Vs(jw). Determine the type of filter (i.e. lowpass, highpass) (20 pts) Vs Rz R₁ L +1 Noarrow_forward1. A balanced three-phase voltage source with a line-to-line magnitude of 208 volts rms supplies a balanced delta-connected load of 12 + j9 ohms per-phase. Determine the rms magnitude of the current that will flow in the lines connecting the voltage source to the load. (20 pts)arrow_forward
- 5. For the circuit shown below determine the following quantities: a. An expression for the transfer relationship H(jw) = Vo(jw) / Vs(jw). b. The maximum value of the transfer relationship. c. Identify the type of filter (i.e. low pass, high pass) d. Determine the cutoff frequency if R₁ = 10k ohms, R2 = 12k ohms and L= 10H. (20 pts) + Vs ① R₁ + L Vo R2arrow_forwardpower systemsarrow_forwardpower systemsarrow_forward
- power systemsarrow_forward1. A three-phase transformer with Yd connection, 300 kVA, 12000/220 V, has been short-circuit tested on the high voltage side giving the following results: 750 V, 14.434 A, 10838 W.When the transformer is connected to nominal voltage without load it consumes 5400W. Calculate:to. Relative voltages of short circuit high voltage side: εcc, εRcc and εXcc.b. The voltage in the secondary when the transformer is connected to nominal voltage and feeds a load of 200 kW fp = 0.8 in delay.c. Apparent power of maximum efficiency and maximum efficiency with fp = 0.95 inductive. solve WITHOUT using artificial intelligence.Solve by hand by one of the collaborators pleasearrow_forward2. A three-phase transformer connection Yy, 2000 kVA, 20000/6000 V has the relative short-circuit voltages Ecc = 7% and ERcc = 1.7%.It is known that when empty this transformer consumes a power Po = 12.24 kW. Calculate:a. Parameters Zcc, Rcc, Xcc, referring to the primary and EXcc.b. If the transformer is connected at rated voltage and feeds a load of 1800 kVA, fp = 0.8, calculate the line voltage at the secondary.c. The maximum apparent power, and the maximum efficiency fp = 0.8 inductive. solve WITHOUT using artificial intelligence.Solve by hand by one of the EXPERTS pleasearrow_forward
- Design a fuel cell stack for a fuel cell bus to operate at 200V and provides 100 HP, (1HP = 750 W). Assume the optimum current density on fuel cell electrode at 1 A/cm2. Thickness of each cell is 0.5cm, and nominal cell voltage is 1V. Calculate the power density of the fuel cell stack. Calculate the voltage drop of the fuel cell stack at 150A if the cell resistance is 2mW. Calculate the required hydrogen fuel (in kg) if the fuel cell operates continuously for 5-hours with 100HP.arrow_forward3. A three-phase Dy connection transformer, 500 kVA, 12000/500 V, has been tested for vacuum on the low voltage side and short circuit on the high voltage side, giving the following results:Vacuum test: Vo = 500 V, Io = 30 A, Po = 900 W.Short circuit test: Vcc = 800 V, Icc = 24.056 A, Pcc = 17233.42 W.Calculate:A. Relative voltages of short circuit high voltage side: Ecc, ERcc and EXcc.B. The voltage in the secondary when the transformer is connected to nominal voltage and feeds a load of 200 kW fp = 0.8 in delay.C. Maximum efficiency with fp = 0.95 inductive. Solve WITHOUT using artificial intelligence.Solve by hand by one of the EXPERTS in the field.arrow_forwardDescribe the advantages and disadvantages of supercapacitor versus battery. Explain the principle operation of Pseudo-capacitor and its advantages -disadvantages versus capacitors.arrow_forward
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