Unformatted text preview: r wave. The
amplitude of the fundamental component of Ip the voltage across the inductor is: T =1/2π A. 4Ip
B. 8Ip
C. 16Ip Ip D. 32Ip
E. None of the above.
Solution: v = L 2I p
di
= 1×
= 8πI p , which is the amplitude of the square waveform
dt
1 / 4π representing v. the amplitude of the fundamental is 4 × 8πI p = 32I p π . 15. The voltage and current at the terminals of a circuit are: v = 15 + 400 cos 500 t + 100 sin 1500 t V + i = 2 + 5 sin(500t + 60 ) + 3 cos(1500t − 15 ) A
o 3% o 3 ∑ Vm I m
1
1
cos(θ vn − θ in ) = 15×2 + × 400 × 5 cos(30 o ) + × 100 × 3 × cos( −75 o )
2
2
2
n =1 = 934.85 W b) Calculate the rms value of v. Vrms = (15)2 + 2% v
– a) Calculate the average power delivered to the circuit. P = Vdc I dc + 3% i ( 400 )2 (100 )2
+
= 291.93 V
2
2 c) Calculate the rms value of i. I rms = (2)2 + (5 ) 2 (3 ) 2
+
= 4.58 A
2
2 2/2 t, s f(t) 12. For n = 1, 2, 3,…, the function shown has:
A. an and bn nonzero for all n
B. an and bn are zero for even n A C. an and bn are zero for odd n
T A D. an = 0 for all n
T/2 E. bn = 0 for all n t T/2 Solution: When the dc value is removed, the ac
component has halfwave symmetry but is neither even nor odd. Hence, an and bn are
zero for even n. 13. Determine the total power dissipated if vI is a fullwave 5Ω rectified waveform given by: vI = 6sin(500t) V.
Solution: Pdc = 2V
1
1
=
= 2 × 10 −5 << 5 ohms; Vdc = m ;
ωC 500 × 100
π +
vI 5Ω 100 F – 2
4Vm
2
= 0.04053Vm ;
2
10π 2
2
2
0.09472Vm
Vm 4Vm
4 ⎞
2
2
2
2⎛ 1
2
0.01894Vm ;
= 2 + Vac ; Vac = Vm ⎜ − 2 ⎟ = 0.09472Vm ; Pac =
5
π
2
⎝2 π ⎠
2
P = 0.05947Vm . 14. A period of a periodic function f(t) is given by: K(4 + f(t) 2sint), 0 < t < 2π. Determine the rms value of f(t), if K =
0.5. ( ) Solution: The square of f(t) is K 2 16 + 16 sin t + 4 sin2 t = K (16 + 2 + 16 sin t − 2 cos 2t ) . The area under the square is
2 ∫ 2π 0 K 2 (16 + 2 + 16 sin t − 2 cos 2t )dt = 36πK 2 ; the mean square is value is 3 2K . 1/2 2
4 π 2π t 36πK 2
= 18K 2 and the rms
2π 16. Derive the trigonometric Fourier expansion of the given f(t) periodic function f(t). 2
1
2 1 1
1 2 2 Solution: Since f(t) is odd, a0 = 0 = an; T = 2, ω0 = 2π/T = π; f(t) = t + 1;
bn = 4 T...
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This test prep was uploaded on 03/11/2014 for the course EECE 210 taught by Professor Riadchedid during the Spring '07 term at American University of Beirut.
 Spring '07
 RiadChedid

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