EBK PRINCIPLES OF ELECTRIC CIRCUITS
10th Edition
ISBN: 9780134880068
Author: Buchla
Publisher: VST
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A three-phase, 400 V system has the following load
connected in delta: between the red and yellow lines,
a non-reactive resistor of 100 Ω; between the yellow
and blue lines, a coil having a reactance of 60 Ω and
negligible resistance; between the blue and red lines,
a loss-free capacitor having a reactance of 130 Ω.
Calculate: (a) the phase currents; (b) the line currents.
Assume the phase sequence to be R–Y, Y–B and
B–R. Also, draw the complete phasor diagram.
ANS:
4.00 A, 6.67 A, 3.08 A, 6.85 A, 10.33 A, 5.8 A
With the aid of a circuit diagram, show that two
wattmeters can be connected to read the total power in
a three-phase, three-wire system.
Two wattmeters connected to read the total power
in a three-phase system supplying a balanced load read
10.5 kW and −2.5 kW respectively. Calculate the total
active power.
Drawing suitable phasor diagrams, explain the
significance of: (a) equal wattmeter readings; (b) a zero
reading on one wattmeter.
ANS:
8 kW
A factory has the following load with power factor of 0.9 lagging in each
phase. Red phase 40 A, yellow phase 50 A and blue phase 60 A. If the supply
is 400 V, three-phase, four-wire, calculate the current in the neutral and the
total active power. Draw a phasor diagram for phase and line quantities.
Assume that, relative to the current in the red phase, the current in the yellow
phase lags by 120° and that in the blue phase leads by 120°.
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.Similar questions
- Fundimentals of Energy Systems Q8arrow_forwardTwo wattmeters are used to measure power in a three phase, three-wire network. Show by means of connection and complexor (phasor) diagrams that the sum of the wattmeter readings will measure the total active power. Two such wattmeters read 120 W and 50 W when connected to measure the active power taken by a balanced three-phase load. Find the power factor of the load. If one wattmeter tends to read in the reverse direction, explain what changes may have occurred in the circuit ANS: 0.815arrow_forwardFundimentals of Energy Systems Q7arrow_forward
- If Va = 12V, Ve = 0V, R1= 10 Ohms, and R2=R3=5 Ohms, solve all currents, i, and voltages, V in the circuit.arrow_forward* 7.29 The current source in the circuit of Fig. P7.29 is given by is(t) = 12 cos(2л × 10¹t — 60°) mA. - Apply the phasor-domain analysis technique to determine ic(t), given that R= 20 2 and C = 1 μF.arrow_forwardPROBLEMS 7.33 Find ia(t) in the circuit of Fig. P7.33, given that Us(t) = 40 sin(200t -20°) V.arrow_forward
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