RMS of square wave and alternating currents

  • Thread starter moenste
  • Start date
  • #1
moenste
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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)?
 

Answers and Replies

  • #2
gneill
Mentor
20,948
2,892
Your 3.45 A for (b) looks good.
 
  • #3
moenste
711
12
Your 3.45 A for (b) looks good.
So it's probably a typo in the book?

And (c) logic is correct?
 

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