11. There is 900 mA of current through a wire with 40 turns. What is the reluctance of the circuit if the flux is 400Wb?
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- 11. There is 900 mA of current through a wire with 40 turns. What is the reluctance of the circuit if the flux is 400Wb? 14,400 At/Wb B 1,440 At/Wb Ⓒ9,000 At/Wb O90,000 At/Wb Answer: Option O Explanation: No answer description is available. Let's discuss.1. What is the reluctance of 10 cm diameter silicon steel of 2500 permeability with a mean length of 20 cm? A. 0.000201 cgs B. 0.000102 cgs C. 0.000212 cgs D. 0.000123 cgs 2. At what rate the flux must be varied to measure 12 V across the terminal of 300 turns coil? A. 0.05 Wb/s B. 0.04 Wb/s C. 0.03 Wb/s D 0.02 Wb/sHow temperature effect conductivity? Explain with the help of following equation. m e'n7'
- show thatvthis is a magnetic flux density: then prove and interpret it with reasons on each stepA magnetic circuit has a length of 100 cm and cross-sectional area of 5 square cm. The total flux is 10 x 104 webers. The coil has 100 turns and current of 4 amperes. What is the magnetic strength? a. 500 AT/m b. 400 AT/m c. 350 AT/m d. 450 AT/mIn 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
- What will be the MMF required to produce a flux density of 0.88 T in an air gap of 8 mm?25 PROBLEMS 12 cm +1.5-mm gap IP1000t Armature (sheet steel) Cast steel A = 2 cm? throughout FIGURE 1.27. The relay shown in Figure 1.27 has a core made of cast steel and an armature of sheet steel. A flux of 0.8 T in the air gap produces a force that pulls in the CO armature. a. Determine the required current to pull in the armature. b. What force is exerted on the armature to overcome the spring tension? c. Assuming in the closed position the air gap is reduced to 0.1 mm, to what value may the current be reduced (dropout current of relay) just to retain closure? Assume that the same force is required in the closed position to overcome the spring force.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
- 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?Problem 1A core with a coil has 240 turns, to generate a flux ø of 0.008482 weber, how much should the current i1 be and what will its direction be? Relative permeabilities: metal 1 of 250 and metal 2 of 375. Core depth 25cmUsing figures (a) and (b), a magnetic flux density of 0.1 Wb in the air gap Determine the current I in the coil with 200 turns required to produce