RMS of square wave and alternating currents

Yes, your logic for (c) is correct. The RMS value of an alternating current is the same as the RMS value of a steady DC current with the same effect.
  • #1
moenste
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12

Homework Statement


Find the value of the RMS current in the following cases:
(a) a sinusoidally varying current with a peak value of 4.0 A,
(b) a square wave current which has a constant value of 4.0 A for the first 3 ms and -2.4 A for the next 2 ms of each 5 ms cycle,
(c) an alternating current which has the same effect as a steady DC current of 2.4 A,
(d) a 240 V RMS supply driving current through a 16 Ω resistor.

Answers: (a) 2.8 A, (b) 3.3 A, (c) 2.4 A, (d), 15 A

2. The attempt at a solution
(a) I = I0 / √2 = 4 / √2 = 2.8 A

(b) Average value of I2 = 42 + 42 + 42 + (-2.4)2 + (-2.4)2 / 5 = 11.904 → RMS = √11.9 = 3.45 A. Did I miss something? I would say a difference of 0.15 / 0.2 is relatively significant, though I don't see any mistakes.

(c) An alternating current which has a same effect as a steady DC current of 2.4 A has an RMS value of 2.4 A.

(d) V = IR → I = V / R = 240 / 16 = 15 A

Are (b) and (c) right? Did I miss something or it's just a typo in (b)?
 
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  • #2
Your 3.45 A for (b) looks good.
 
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  • #3
gneill said:
Your 3.45 A for (b) looks good.
So it's probably a typo in the book?

And (c) logic is correct?
 

Related to RMS of square wave and alternating currents

1. What is the RMS of a square wave?

The root mean square (RMS) of a square wave is the equivalent steady direct current (DC) value that would produce the same amount of heat as the square wave in a given resistor. It is calculated by taking the square root of the average of the square of the signal's amplitude over one period.

2. How do you calculate the RMS of a square wave?

To calculate the RMS of a square wave, first find the amplitude of the wave by measuring the vertical distance between the highest and lowest points on the wave. Then, square this value and divide it by the period of the wave. Finally, take the square root of this result to find the RMS value.

3. What is the difference between RMS and peak value in a square wave?

The peak value of a square wave is the maximum amplitude of the wave, while the RMS value is the equivalent steady DC value that would produce the same amount of heat. The RMS value takes into account the changing amplitude of the square wave, while the peak value only considers the highest point on the wave.

4. What is the RMS of an alternating current (AC) signal?

The RMS of an AC signal is the equivalent steady DC value that would produce the same amount of heat as the AC signal in a given resistor. It is calculated by taking the square root of the average of the square of the signal's amplitude over one full cycle.

5. How does the RMS of an AC signal compare to its peak value?

The RMS value of an AC signal is always lower than its peak value. This is because the RMS value takes into account the changing amplitude of the signal, while the peak value only considers the highest point on the wave.

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