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Name___________________________________
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Asinusoidal wave form hasthe expression v(t) = 20sin(1000t). The cyclic frequency is
A 1 kresistor is in series with a 3 kresistor. Agenerator with asignal output of 24sin(2000t) Vis
connected to this circuit. The voltage across the 1 kresistor is
Using the effective value of an AC voltage results in
half the average power as aDC source of the same value.
the same average power as aDC source of the same value.
twice the average power as aDC source of the same value.
Asinusoidal signal is called “steady–state” if
the peak amplitude is constant.
the frequency is constant.
Asinusoidal wave form hasthe expression v(t) = 20sin(1000t). The RMS magnitude is
Asinusoidal wave form hasthe expression v(t) = 20sin(1000t). The peak value is
A20 resistor is in parallel with a40 resistor. That combination is in series with a60 resistor.
A12 VRMS AC source is connected to the circuit. The voltage across the 60 resistor is
A 1 kresistor is in series with a 3 kresistor. A24sin(2000t) Vsource is connected to this circuit.
The current through the 1 kresistor is
You measured v(t) = 0.220sin(20000t +
4)across a470resistor. The RMS current is
A32 resistor is in parallel with a74 resistor. The total current is 13 mARMS.The current
through the 74 resistor is
You measured v(t) = 10cos(200t – 40°) across a 2 kresistor. The peak current is
The average power in an AC resistive circuit is
peak current squared, times the resistance.
RMS current squared, divided by the resistance.
peak voltage times peak current.
RMS voltage squared, divided by the resistance.
Anegative phase shift angle in the expression for asinusoidal signal crosses the horizontal axis
(going from negative to positive values)
A10 kresistor is connected in parallel with a100kresistor. A 10sin(10000t) Vsource is applied.
The current through the 100kresistor is
Asinusoidal wave form hasthe expression v(t) = 20sin(1000t). The period is
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
Given i(t) = 20sin(200t + 20°), i(2 ms) = ________.
Given a VP = 50 Vand an IRMS = 100 µA, the average power is ________.
You measure 10 Vpeak–to–peak on the oscilloscope across a200resistor. The power
dissipated by the resistor is ________.
Given v(t) = 100sin(2000t + 45°),the time that the peak of the signal occurs is at ________.
Given a VRMS = 25 Vand an IP = 4.5 mA, the average power is ________.
A 1 , 2 ,and 3resistor are connected in parallel. A 5 mA AC current source is applied
to the circuit. The power dissipated by the 2resistor is ________.
A 1 , 2 ,and 3resistor are connected in parallel. A 5 mA AC current source is applied
to the circuit. The current through the 2resistor is ________.
Given i(t) = 100sin(2000t + 20°),i(10 ms) = ________.
Given a VP = 10 Vand an IP = 2mA, the average power is ________.
Given i(t) = 20sin(200t + 20°), i(10 ms) = ________.
Given i(t) = 20sin(200t + 20°), i(1 ms) = ________.
Given v(t) = 100sin(2000t + 45°),v(1 ms) = ________.
You measure 10 VAC on aDMM across a 1 kresistor. The power dissipated is ________.
Aphase shift of
3is ________ degrees.
Aphase shift of 222°is ________ radians.
TRUE/FALSE. Write ‘T’ if the statement is true and ‘F’ if the statement is false.
Anegative phase shift advances the signal in time. The waveform is shifted to the left.
Apositive phase shift advances the signal in time. The waveform is shifted to the right.
Apositive phase shift advances the signal in time. The waveform is shifted to the left.
The peak value is 1.414 times the RMS value.
Anegative phase shift delays the signal in time. The waveform is shifted to the right.
An AC signal hasaconstant voltage over time.
The average power is the RMS voltage times the RMS current.
The RMS value is 1.414 times the peak value.
DC series–parallel techniques can be applied to AC series–parallel resistive circuits.
The average power is the peak voltage times the peak current.
The RMS value is 0.707 times the peak value.
Asinusoidal waveform expression is v(t) = VPsin( t).
Electrical power transmission can be very efficient with AC.
DC series–parallel techniques will not work with AC signal sources.
An AC signal can be represented by asinusoidal waveform.