Physics · Alternating Current · NEET
If you average I_0 sin(wt) over one full cycle you get zero, because the positive and negative halves cancel. That zero tells you nothing about heating. So we square the current first. Squaring makes every value positive (a negative current heats a resistor just as much as a positive one), so the average of the squared current is not zero. We then take the square root at the end to bring the units back to amperes. This 'square, average, root' order is exactly what root-mean-square means.
It comes from one fact: the average of sin squared over a full cycle is 1/2. Write i = I_0 sin(wt). Then i squared = I_0 squared sin squared(wt). The average of sin squared(wt) over one cycle is 1/2 (proved using sin squared = (1 - cos2wt)/2, and cos2wt averages to zero). So mean of i squared = I_0 squared x (1/2). Take the square root: I_rms = I_0 x root(1/2) = I_0/root2 = 0.707 I_0.
Yes, exactly. Use the identity sin squared(wt) = (1 - cos2wt)/2. Over one full cycle the average of cos2wt is zero (it is a cosine, equal positive and negative area). So the average of sin squared = (1 - 0)/2 = 1/2. The same is true for cos squared. This 1/2 is the whole reason the factor is 1/root2 and not something else.
Yes. The derivation only uses the shape of the sine wave, not whether it is current or voltage. So V_rms = V_0/root2 = 0.707 V_0 exactly the same way. NEET often asks it as V_0 = root2 V_rms (the peak is root2 times the rms). Both statements are the same equation rearranged.
RMS is the value that produces the same heating (same average power) in a resistor as a steady DC current of that size. When we say Indian mains is 220 V, that 220 V is the RMS value, not the peak. The peak is actually 220 x root2 = about 311 V. All AC power and meter readings use RMS, which is why the 0.707 factor keeps appearing in numerical problems.
The peak voltage of the ac source is equal to:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
0.707 is the decimal value of 1/root2. Since root2 is about 1.414, 1 divided by 1.414 is about 0.707. So I_rms = I_0/root2 = 0.707 I_0.
RMS is always smaller than the peak for a sine wave. I_rms = 0.707 I_0, so the RMS is about 70.7 percent of the peak. The peak is the largest instantaneous value; the RMS is an effective average, so it must be less.
No. The factor 1/root2 = 0.707 is only for a pure sine (or cosine) AC. For a square wave RMS = peak, and for a triangular wave RMS = peak/root3. NEET's syllabus deals with sinusoidal AC, so 0.707 is the value you use.
The 220 V mains is the RMS value. Its peak is V_0 = root2 x 220 = about 311 V. This is a classic NEET check that you know which number is peak and which is RMS.
Because an AC of RMS value I_rms produces exactly the same heating (same average power I squared R) in a resistor as a steady DC current of the same numerical value. That equal-heating property is why RMS is the 'effective' value.
