1.28 The coils of the magnetic circuit shown in Fig. 1.36 are connected in series so that the mmf's of paths A and B both tend to set up flux in the center leg C in the same direction. The coils are wound with equal turns, \( \mathrm{Nl}=\mathrm{N} 2=100 \).
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In this case, the coils are connected in series, which means the same current flows through both of them. Second, since the coils are wound with equal turns, the mmf of each path (A and B) will be the same. Third, since the mmf's of paths A and B both tend to Show more…
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Problem 2: [35 points] A two-legged core is shown in Figure 2. The winding on the left leg of the core (N1) has 400 turns, and the winding on the right (N2) has 200 turns. The coils are wound in the directions shown in the figure. The depth of the core is 10 cm. If the dimensions are as shown, then what flux would be produced by currents I1=0.5 A and I2=1.0 A? Assume μr=1000 and constant. (Hint: Two mmfs will add up)
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The magnetic circuit in Figure 1 has cast steel core with dimensions as shown. The mean path length from A to B through either outer leg is 0.5 m. The mean path length from A to B through the central leg is 0.2 m. It is required to establish a flux of 0.75 mWb in the air gap of the central leg. Determine the MMF of the coil if the core material has infinite permeability. The depth of the core is 1cm, and the left, top, right and bottom legs are 1cm in width while the center leg is 2cm in width. Ignore fringing effects. (μ₀ = 4π x 10⁻⁷ H/m)
1.32 The magnetic circuit of Fig. 1.35 has two windings and two air gaps. The core can be assumed to be of infinite permeability. The core dimensions are indicated in the figure. a. Assuming coil 1 to be carrying a current I1 and the current in coil 2 to be zero, calculate (i) the magnetic flux density in each of the air gaps, (ii) the flux linkage of winding 1, and (iii) the flux linkage of winding 2. b. Repeat part (a), assuming zero current in winding 1 and a current I2 in winding 2. c. Repeat part (a), assuming the current in winding 1 to be I1 and the current in winding 2 to be I2.
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