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ECE286 Final Exam MCQ 2024 QUESTIONS & ANSWERS SOLVED 100- CORRECT_.pdf
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8/14/24, 4:53 AMECE286 Final Exam MCQ
Jeremiah
Terms in this set (63)
Which of the following is not a right way to A * sin(wt - 90°)
express the sinusoid A * Cos(wt)
a.) A * cos(2πft)
b.) A * cos(2πt/T)
c.) A * cos(w(t-T))
d.) A * sin(wt - 90°)
A function that repeats itself after fixed periodic
interval is said to be:
a.) a phasor
b.) harmonic
c.) periodic
d.) reactive
1k Hz
Which of these frequencies has the shorter w = 2πf
period? f = w/2π
a.) 1k rad/s Fa = 1000/2π = 159.15
b.) 1 kHz Period a = 1/159.15 = 6.28x10^-3
Period b = 1/1000 = 1x10^-3
If V1 = 30sin(wt + 10°) and V2 = 20sin(wt + 50°), v2 leads v1
which of these statements are true?
a.) v1 leads v2 v1 lags v2
b.) v2 leads v1
c.) v2 lags v1
d.) v1 lags v2
c.) v1 and v2 are in phase
The voltage across an inductor leads the True
current through it by 90° (T/F)
The imaginary part of impedance is called Reactance
a.) Resistance
b.) Admittance
c.) Susceptance
ECE286 Final Exam MCQ
The impedance of a capacitor increases with Xc = 1/( jwC)
increasing frequency (T/F) False because w is in the denominator
A series RC circuit has |VR| = 12V and |VC| = √ |12|^2 + |5|^2 = √144+ 25 = √169 = 13V
5V, the magnitude of the supply voltage is:
a.) -7V 13V
b.) 7V
c.) 13V
d.) 17V
d.) Conductance
e.) reactance
1/8
, 8/14/24, 4:53 AM
A series RCL circuit has R = 30Ω, Xc = 50Ω, Xc is negative
and XL = 90Ω. The impedance of the circuit XL is positive
is:
a.) 30 + j140Ω 30 -jXc + jXL
b.) 30 + j40Ω 30 - 50j + 90j = 30 + 40j
c.) 30 - j40Ω
d.) -30 - j40Ω
e.) -30 + j40Ω
Pspice can handle a circuit with two False
independent sources of different
frequencies
(a) True
(b) False
The average power absorbed by an inductor True
is zero (T/F)
The Thevenin impedance of a network seen Maximum power transfer theorem: Maximum power will be transferred from source to
from the load terminals is 80 + j55Ω. For load when the load impedance is equal to complex conjugate of source impedance
maximum power transfer, the load
impedance must be: C
a.) -80 + j55Ω
b.) -80 - j55Ω
c.) 80 - j55Ω
d.) 80 + j55Ω
The amplitude of the voltage available in the Vout = Vin / √2
60-Hz, 120-V power outlet in your home is: 120 = x /√2
a.) 110 V x = 120 * √2 = 170V
b.) 120 V
c.) 170 V C
d.) 210 V
If the load impedance is 20-j20, the power Pf = P/S = cos(θv - θi)
factor is Z = Vrms/Irms = Vrms/Irms * ∠θv - θi 20
a.) ∠-45° -j20 = 28.28∠-45°
b.) 0 cos(-45°) = 0.7071
c.) 1
d.) 0.7071
e.) none of the above
ECE286 Final Exam MCQ
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