Solution the voltage across all windings is zero

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Unformatted text preview: .5 Ω. Supply voltage -j30 Ω L Solution: The reactive power absorbed by the load is V2 1200 = −3000 × 0.6 = 900 VAR. The reactive power absorbed by the capacitor is − 30 0.8 VAR. The total complex power is 1200 + j(900 – 3000) = 1200 – j2100 VA. The magnitude of the line current (1200)2 + (2100)2 = 65 A. The power dissipated in Rline is 65Rline. 300 9. Two identical coils, each having an inductance of 10 mH, are connected in series. When the connections to one of the coils are reversed, the total inductance is multiplied by a factor a. Determine the coupling coefficient of the coils. Solution: (10 + 10 + 2M) = a(10 + 10 – 2M); 2M(a + 1) = 20(a – 1); M= 2Ω R 10(a − 1) M (a − 1) ; k= = a +1 10 a +1 100 turns 10. Determine Ix, assuming R = 4 Ω. Solution: The voltage across all windings is zero. Hence, I1 = I2 = 10∠0° V + – 4∠0° V – Ix 400 turns 10 A, and R + 200 turns 4 = 2 A. Setting the net mmf to 2 R 2Ω I1 I2 100 turns zero, 400I1 – 100I2 + 200I3 = 0, or 4 × 10 − 2 + 2 I3 = 0, which gives I3 = R 20 20 ; IX = I2 – I3 = 1 + . 1− R R 10∠0° V + – Ix 400 turns I3 4/5 200 turns + – 4∠0° V 1. If M = 5 mH, determine the ratio v1/v. Solution: Leq = L1 + L2 – 2M; v = Leq hence, M di di di ; v 1 = L1 ; −M dt dt dt i v1 L1 − M 60 − M = = v L1 + L2 − 2M 100 − 2M + + 60 mH 40 mH _ v1 2. Determine VO, given that I = 1∠0° A and ω = 10 rad/s. k= 0.5 Solution: M = k L1L2 = 0.5 H; secondary voltage is jωMI, + 1H I with the dotted terminal positive with respect to the _ v 1H undotted terminal. Hence, VO = -jωMI = -j10×0.5I = -j5I. VO _ 3. The lamp glows brighter when the dots are at coil terminals 1 Solution: The lamp glows brighter when the voltage across it is largest. This occurs when the voltages across the windings are 1′ additive, that is, when the dots are at terminals 1 and 2 or 1′ and 2′. 2 12 V ac 100 turns 200 turns 2′ 5Ω I 4. Determine the reactive power absorbed in the circuit, given that I = 1∠0° A rms. j5 Ω Solution: The equivalent series impedance is -j5 Ω j 5(5 − j 5 ) = 5 + j 5 . The reactive power is 5|I|2 VAR. As a j5 + 5 − j5 check, the current in the capacitive branch is j...
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