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).
Introductory Circuit Analysis (13th Edition)
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Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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Transcribed Image Text:An iron circuit with a small 0.75 mm air gap is shown in Figure 1. A 6000 turn coil carries a current
1 = 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 (t, = 800 for iron).
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