or
Subelement B
Electrical Math
Section 12
Waveforms
At pi/3 radians, what is the amplitude of a sine-wave having a peak value of 5 volts?
• -4.3 volts.
• -2.5 volts.
• +2.5 volts.
+4.3 volts.

At $\pi$/3 radians, what is the amplitude of a sine-wave having a peak value of 5 volts?

** +4.3 volts.**

$\pi$ 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?
• -4.3 volts.
• -2.5 volts.
+2.5 volts.
• +4.3 volts.

At 150 degrees, what is the amplitude of a sine-wave having a peak value of 5 volts?

+2.5 volts

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.
• -2.5 volts.
• +2.5 volts.
• +4.3 volts.

At 240 degrees, what is the amplitude of a sine-wave having a peak value of 5 volts?

-4.3 volts.

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?
• AC voltage is the square root of the average AC value.
• The DC voltage causing the same heating in a given resistor at the peak AC voltage.
• The AC voltage found by taking the square of the average value of the peak AC voltage.
The DC voltage causing the same heating in a given resistor as the RMS AC voltage of the same value.

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.

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?
• 170 volts AC.
• 240 volts AC.
120 volts AC.
• 350 volts AC.

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.
• A is lagging B by 90 degrees.
B is lagging A by 90 degrees.
• A is leading B by 180 degrees.
• B is leading A by 90 degrees.

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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