Problems_after_Quiz_3

The amplitude of the fundamental is 4 8i p 32i p 15

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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 half-wave 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 full-wave 5Ω rectified waveform given by: vI = 6|sin(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.

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