The RMS (root mean square) value of an alternating current is the steady DC current that would produce the same heating in a resistor over one full cycle. For a sine wave, I_rms = I0/√2 = 0.707 I0 and V_rms = V0/√2, where I0 and V0 are the peak values. Memory hook: RMS is the "DC-equal" or "effective" value, so when your home supply says 220 V, that number is already the RMS voltage, not the peak.
A sine AC wave with peak value I0 (green line), RMS value I_rms = 0.707 I0 (red dashed line, the effective DC-equal level), and average over a full cycle equal to zero. RMS sits at about 71% of the peak because heating depends on the average of i², not on i itself.
Your doubts, answered
Is the 220 V house supply the RMS value or the peak value?
It is the RMS value. When any AC supply is rated, the number given is always the RMS value because that is what causes the real heating and work. So 220 V is V_rms. The peak (maximum) voltage is actually higher: V0 = √2 × 220 ≈ 311 V. Many students wrongly plug 220 into peak formulas. Remember: rated AC value = RMS value.
Why do we use the RMS value instead of the simple average value?
Over one full cycle the AC current spends equal time positive and negative, so its plain average is zero. A zero average tells you nothing useful about heating. But heating depends on i squared (power = i²R), and i² is always positive. So we square the current, take its mean (average), then take the square root, giving the RMS value. This RMS value is non-zero and correctly measures the effective heating current.
What does the RMS current physically measure?
It measures the equivalent DC. If an AC current has I_rms = 2 A, it heats a resistor exactly like a steady 2 A DC current would. That is why RMS is also called the effective or virtual value. This is the whole reason it is defined this way: heat produced (H = I_rms² R t) matches the DC case.
Is the relation I_rms = I0/√2 true for every waveform?
No. The factor 1/√2 (= 0.707) only holds for a pure sine (or cosine) wave. For a square wave, triangular wave, or half-wave rectified current, the RMS-to-peak ratio is different. In NEET, the AC source is almost always sinusoidal, so I_rms = I0/√2 applies, but do not blindly use it if the waveform is not a sine.
Are the RMS relations the same for voltage and current?
Yes. For a sinusoidal source, both follow the identical rule: V_rms = V0/√2 and I_rms = I0/√2. The peak is always √2 times the RMS for both. So if a NEET question gives you peak voltage, divide by √2 to get RMS; if it gives RMS, multiply by √2 to get peak.
⚠️ The NEET trap ✗ Using V0 = 220 V (treating the rated supply voltage as the peak value). ✓ The rated 220 V is the RMS value; the peak is V0 = √2 × V_rms = √2 × 220 ≈ 311 V. 🧠 Rated AC value is ALWAYS RMS. Peak = √2 × rated. Multiply, do not divide, to get the peak.
Real NEET questions
NEET 2022
The peak voltage of the ac source is equal to:
A · the value of voltage supplied to the circuit
B · the rms value of the ac source
C · √2 times the rms value of the ac source ✓
D · 1/√2 times the rms value of the ac source
Solution: The RMS value is defined by V_rms = V0/√2. Rearranging for the peak: V0 = √2 × V_rms. So the peak voltage equals √2 times the RMS value of the AC source. Option (A) is wrong because the value 'supplied' or rated is the RMS value, not the peak. Option (D) has the ratio inverted. Correct answer: (C).
NEET 2020
A 40 μF capacitor is connected to a 200 V, 50 Hz ac supply. The rms value of the current in the circuit is, nearly:
A · 2.5 A ✓
B · 25.1 A
C · 1.7 A
D · 2.05 A
Solution: Step 1: The rated 200 V is V_rms. Step 2: ω = 2πf = 2π(50) = 100π rad/s. Step 3: Capacitive reactance X_C = 1/(ωC) = 1/(100π × 40×10⁻⁶) ≈ 79.6 Ω. Step 4: Because V_rms and X_C are both RMS-based, RMS current I_rms = V_rms/X_C = 200/79.6 ≈ 2.5 A. This shows RMS voltage gives RMS current directly. Correct answer: (A).
Solved Alternating Current NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
For a sinusoidal current, I_rms = I0/√2 = 0.707 I0, and for voltage V_rms = V0/√2 = 0.707 V0, where I0 and V0 are the peak values.
Why is RMS also called the effective value?
Because an AC current of RMS value I_rms produces exactly the same heat in a resistor as a steady DC current of the same value I_rms. It is the DC-equivalent, so it is called the effective or virtual value.
What is the RMS value of a 311 V peak supply?
V_rms = V0/√2 = 311/√2 ≈ 220 V. This is why Indian mains, with a peak of about 311 V, is rated as 220 V RMS.
Does an AC ammeter read RMS or peak current?
An AC ammeter (and voltmeter) reads the RMS value, because it responds to the heating effect (i²R), which is what RMS represents.
Is RMS value greater or smaller than peak value?
It is always smaller. RMS = 0.707 × peak for a sine wave, so the RMS value is about 71% of the peak value.
