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 5I2 VAR. As a
j5 + 5 − j5 check, the current in the capacitive branch is j...
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 Spring '07
 RiadChedid
 Power, Power factor

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