At \(\pi\)/3 radians, what is the amplitude of a sine-wave having a peak value of 5 volts?
** +4.3 volts.**
From amchaddad:
\(\pi\) radians is 180° degrees through the cycle.
\(\pi\)/3 radians is 60° degrees through the cycle, which is about 2/3 of the way to the top ( \(\pi\)/2) of the +5V cycle, making it closest to +4.3 volts.
What we are looking for is instantaneous voltage of a sine wave at a specified angle.
The equation is:
"Instantaneous = peak ∗ sin(angle)"
In this case it is answer = 5 ∗ sin(60°) since the peak voltage is 5 and the angle is 60° degrees.
5 ∗ sin(60°) = 4.33v
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At 150 degrees, what is the amplitude of a sine-wave having a peak value of 5 volts?
+2.5 volts
From amchaddad.:
Logically, 150° degrees is still on the +V part of the cycle, so eliminate the negative volt choices. Negative voltage starts at 180° degrees.
Now, +4.3 volts is near the +5 volt peak (closer to 120° degrees) , which is too high for this point of the cycle. 150° degrees is 1/2 peak voltage, or +2.5 volts.
The formula is for the instantaneous voltage
Y = Vp ∗ sin(∢ angle)
So the answer in this case is
5 ∗ sin(150°) = 2.5Volts.
Use the sin ▣ button/function on most scientific calculators. Easy.
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At 240 degrees, what is the amplitude of a sine-wave having a peak value of 5 volts?
-4.3 volts.
From amchaddad.:
240° degrees is on the negative voltage phase of the cycle (180° degrees to 360° degrees).
So, eliminate the positive voltage answers.
The choice is now between −2.5V and −4.3V.
The 240° degrees is 30° degrees off of peak −5V, so we are still closer to that peak voltage in the cycle, making −4.3 volts the correct answer.
(−2.5 volts is reached at 210° and 300° degrees.)
The formula is instantaneous voltage
Y = Vp ∗ sin(∢ angle)**
So, the answer in this case is
5 ∗ sin(240°) = −4.3Volts.
Use the sin ▣ button/function on most scientific calculators. Easy.
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What is the equivalent to the root-mean-square value of an AC voltage?
The DC voltage causing the same heating in a given resistor as the RMS AC voltage of the same value.
For more information, please see Electronics Tutorials site for the article RMS Voltage Tutorial
Also, see Electrical 4 U site for the article on RMS Voltage: What it is? (Formula And How To Calculate It)
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What is the RMS value of a 340-volt peak-to-peak pure sine wave?
120 volts AC.
From wp2ahg:
Starting with 340 volts peak-to-peak. Divide by 2 to get PEP:
\[\frac{340\text{ Volts}}{2} = 170\text{ Volts}\]
Multiply by 0.707 to get RMS volts AC:
\[170\text{ volts} ∗ 0.707 = 120\text{ volts AC}\]
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Determine the phase relationship between the two signals shown in Figure 3B3.
B is lagging A by 90 degrees.
From B to A is 90° degrees. Because it is to the left, it is actually -90° degrees, or lagging.
For well-illustrated explanation, please see Electronics Tutorials site for the article Phase Difference and Phase Shift,
especially section on Phase Difference between a Sine wave and a Cosine wave
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