An iron circuit with a small 0.75 mm air gap is shown in Figure 1. A 6000 turn coil carries a current I= 18 mA which sets up a flux within the iron and across the air gap. The cross section of the iron is a consistent 0.8 cm², and the mean length of the flux path is 0.15 m. a) Redraw the magnetic circuit using schematic symbols of an electric circuit with reluctance in each part of the circuit. b) State's Ohm's Law for magnetic circuit. c) By neglecting the effect of fringing, calculate the reluctance of the circuit. d) Find the flux within the core. N = 6000 Iron circuit (4, = 800 for iron).

An iron circuit with a small 0.75 mm air gap is shown in Figure 1. A 6000 turn coil carries a current I= 18 mA which sets up a flux within the iron and across the air gap. The cross section of the iron is a consistent 0.8 cm², and the mean length of the flux path is 0.15 m. a) Redraw the magnetic circuit using schematic symbols of an electric circuit with reluctance in each part of the circuit. b) State's Ohm's Law for magnetic circuit. c) By neglecting the effect of fringing, calculate the reluctance of the circuit. d) Find the flux within the core. N = 6000 Iron circuit (4, = 800 for iron).

Electricity for Refrigeration, Heating, and Air Conditioning (MindTap Course List)

10th Edition

ISBN:9781337399128

Author:Russell E. Smith

Publisher:Russell E. Smith

Chapter8: Basic Electric Motors

Section: Chapter Questions

Problem 3RQ

See similar textbooks

Similar questions

An iron circuit with a small 0.75 mm air gap is shown in Figure 1. A 6000 turn coil carries a current I = 18 mA which sets up a flux within the iron and across the air gap. The cross section of the iron is a consistent 0.8 cm², and the mean length of the flux path is 0.15 m. a) Redraw the magnetic circuit using schematic symbols of an electric circuit with reluctance in each part of the circuit. b) State's Ohm's Law for magnetic circuit. c) By neglecting the effect of fringing, calculate the reluctance of the circuit. d) Find the flux within the core. N = 6000 Iron circuit (ur = 800 for iron).
An iron circuit with a small 0.75 mm air gap is shown in Figure 1. A 6000 turn coil carries a current I = 18 mA which sets up a flux within the iron and across the air gap. The cross section of the iron is a consistent 0.8 cm2, and the mean length of the flux path is 0.15 m. a) Redraw the magnetic circuit using schematic symbols of an electric circuit with reluctance in each part of the circuit. b) State's Ohm's Law for magnetic circuit. c) By neglecting the effect of fringing, calculate the reluctance of the circuit. d) Find the flux within the core. N = 6000 Iron circuit (4, = 800 for iron).
An iron circuit with a small0.75 mm air gap is shown in Figure 1. A 6000 turn coil carries a current I = 18 mA which sets up a flux within the iron and across the air gap. The cross section of the iron is a consistent 0.8 cm2, and the mean length of the flux path is 0.15 m. a) Redraw the magnetic circuit using schematic symbols of an electric circuit with reluctance in each part of the circuit. b) State's Ohm's Law for magnetic circuit. c) By neglecting the effect of fringing, calculate the reluctance of the circuit. d) Find the flux within the core. N = 6000 Iron circuit (u, = 800 for iron). Figure 1
An iron circuit with a small 0.75 mm air gap is shown in Figure 1. A 6000 turn coil carries a current I = 18 mA which sets up a flux within the iron and across the air gap. The cross section of the iron is a consistent 0.8 cm?, and the mean length of the flux path is 0.15 m. a) Redraw the magnetic circuit using schematic symbols of an electric circuit with reluctance in each part of the circuit. b) State's Ohm's Law for magnetic circuit. c) By neglecting the effect of fringing, calculate the reluctance of the circuit. d) Find the flux within the core. N = 6000 Iron circuit (4, = 800 for iron). Figure 1
Please answer a, b, c and d
The length of a magnetic circuit in a moving iron instrument is 300mm. The coil around the soft iron core has 360 turns and takes a current of 1.75 A. The core is square in section with sides of 20mm. Take the relative permeability of soft iron as 1100. Determine the total flux
In a certain cast-steel core series magnetic circuit with a 400 turns, 170 At/m, mean length of 0.16 m and 0. 002 square meters cross-sectional area. the value of current to develop a 0.0004 Wb is
show thatvthis is a magnetic flux density: then prove and interpret it with reasons on each step
A toroidal core with a mean circumference of 100 cm and a cross-sectional area of 10 cm2 is wound with 500 turns of wire. What current would be required to generate a flux of 1 mWb in the core. Assume the core has a relative permeability of 800
A current in a 250 turn coil is 2.25A, wound in a core that has a reluctance of 12,000 AT/Wb. What is the resulting flux density (in Tesla), if the area of the core is 8cm²?
1.Consider the magnetic core below.All dimensions are shown on the figure. The core is made up of legs with 2 different cross-sectional areas.The depth of core is 10cm.The relative permeability of the core is 2000. The current in the magnetising coil is 1 A. The number of turns of the coil is 2000.Calculate: (i)the reluctances and draw the magnetic circuit (ii)the flux in the core (iii)Flux densities B and field intensities H in legs having different cross-sectional areas. (iv)Verify Ampere's Law. Depth: 10 cm I=1A 20cm 2000 Turns 65cm 15cm 15cm 50cm 15cm ndo un of
The length of a magnetic circuit in a moving iron instrument is 300mm. The coil around the soft-iron core has 360 turns and takes a current of 1.75 A. The core is square in section with sides of 20 mm. Takethe relative permeability of soft iron as 1100. Determine the flux density