EBK PRINCIPLES OF ELECTRIC CIRCUITS
10th Edition
ISBN: 9780134880068
Author: Buchla
Publisher: VST
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a) An iron ring, having a mean circumference of 250 mm and a cross-sectional area of 400
mm², is wound with a coil of 70 turns. Using the following data, calculate the current
required to set up a flux of 510µWb in the ring.
H (A/m) 350
600
1250
B (T)
1.0
1.2
1.4
b) Calculate also:
i. The inductance of the coil at the current obtained in Question 2 (a) above.
ii. The self-induced e.m.f. if this current is switched off in 0.005 s. Assume that there
is no residual flux.
A balanced three-phase, 1351-V, 60-Hz, A-connected source feeds a balanced Y-
connected load with a per-phase impedance of 360 + j150 Q as shown in Figure 1.
Calculate:
i. The readings on each of the wattmeters
ii. The power factor of the load using the wattmeter readings.
NOTE:
i.
ii.
Let VAN be the reference phasor, and the phase sequence be ABC
anticlockwise.
Assume the voltage-drop on the conductors between the source and the load
to be zero volts.
V b
V₁
W
000
000 ;
A
360 + j150
360 + j150
4
b
0000
000
B
360 + j150
C
W₂
Figure 1
a) Three 30 2 resistors are arranged as shown in Figure 1 below. They are connected to
a 480 V three-phase supply. The phase sequence is RYB anticlockwise. Calculate:
i. The total power drawn by the circuit using the phase parameters.
ii.
The power read by each wattmeter.
b) If Za, one of the 30 2 resistors, is now removed from the circuit, calculate:
R-
i.
The line currents: IR, Iv, and la
ii.
The power read by each wattmeter.
iii.
The total power drawn by the two resistors.
W₁
Be-
W2
www
'R
22
12
B
Figure 1
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- A certain magnetic circuit may be regarded as consisting of three parts, A, B and C in series, each one of which has a uniform cross-sectional area. Part A has a length of 300 mm and a cross- sectional area of 450 mm². Part B has a length of 120 mm and a cross-sectional area of 300 mm². Part C is an airgap 1.0 mm in length and of cross-sectional area 350 mm². The flux in the airgap is 0.35 mWb. Neglect magnetic leakage and fringing. The magnetic characteristic for parts A and B is given by: H (A/m) 400 560 800 1280 1800 B (T) 0.7 0.85 1.0 1.15 1.25 Calculate: i. The reluctance of each part, that is, of Part A, Part B, and Part C ii. The total reluctance of the magnetic circuit. iii. The total m.m.f.arrow_forwardHW_02.pdf EE 213-01 Assignments HW_#2 Toms are as muIcate uah.instructure.com b Answered: HW_#1 HW_01.pdf EE 213-01 Assignments P Pearson MyLab and Mastering uah.instructure.com P Course Home Watc... ✓ Download → Info × Close 1) (5 pts)For the circuit shown, find the value of Ia, Ib, Ic and Va: Ib 10 Ohms + Ic 40 Ohms 20 Ohms 70 Volts a Page 1 > of 2 - ZOOM + pui via Canvas Hint: use KCL to find Ic in terms of la and lb, use KVL around the right loop to find a relationship between la and lb, use KVL around the outer loop to solve for a current value (use ohms law for the voltage drops across the resistors) 2) (5 pts) – For problem 1, show that the power supplied (delivered) by the source equals the power absorbed by the resistors. 3) (5 pts) Determine the equivalent resistance looking into terminals a-b for the two circuits shown. Use series and parallel resistor combinations. Hint: work from the opposite end of terminals a-b towards terminals a-b a) 24 Ohms 10 Ohms 8 Ohms G a REQ b)…arrow_forwardA three-phase load consists of three identical impedances connected in delta. The impedances have a value of (100 + j40) Q. The supply voltage is 440 V. The two- wattmeter method is used to determine the total power drawn by the load. One wattmeter, W₁, has its current coil connected in the blue line and its voltage coil connected between the yellow and blue lines. Calculate: a) The power measured by each wattmeter b) The power factor of the load using the wattmeter readings c) The apparent power supplied to the load d) The reactive power drawn by the load.arrow_forward
- 4- A 75 KW, 250 V shunt generator has R₁ = 0.03 2 and RF =50 2. The generator has 6 poles and is lap wound with totally 400 conductors. The generator runs at 600 rpm on full load. The bore of the machine is 42 cm (in diameter), its axial length is 28 cm and each pole subtends an angle of 54 °. Field pole a) b) c) Find the airgap flux density. (10 pt) Find the induced voltage under full load. (10 pt) Find the average number of active conductors. (5 pts) 28 cm 21 cm 60° Armaturearrow_forwarda) The phase currents in a delta - connected three-phase load are as follows: between the red and yellow lines, 60 A at p.f. 0.9 leading; between the yellow and blue lines, 40 A at 0.95 p.f lagging; between the blue and red lines, 50 A at p.f. 0.8 lagging. The supply voltage is 240 V. Draw a labeled diagram of the circuit and calculate the line currents. Note: Use VRY as the reference phasor. b) The two-wattmeter method is used to measure the total power drawn by the load in 1 (a) above. Wattmeter one has its current coil connected in the yellow line and its voltage coil connected between the yellow and red lines. Calculate: i. The power measured by each wattmeter. ii. The power factor of the load.arrow_forwardQUESTION NO. 1 a) State Lenz Law b) Explain Flux density c) A wire, 100 mm long, is moved at a uniform speed of 4 m/s at right angles to its length and to a uniform magnetic field. Calculate the density of the field if the e.m.f. generated in the wire is 0.15 V. If the wire forms part of a closed circuit having a total resistance of 0.04 2, calculate the force on the wire in newtons.arrow_forward
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