When you push a DC current through a resistor, the power is constant — P=I2R, and the heating is steady. But an AC current keeps changing direction and magnitude. At one instant it's +I0, a moment later it's zero, then −I0. If you simply averaged the current over time, you'd get zero — because the positive and negative halves cancel. That's useless for telling you how much heat the resistor actually feels.
So we need a single number that captures the effective heating power of an alternating current. That number is the RMS value.
The Intuition: Squaring Fixes the Sign Problem
Heat depends on I2, not on I. Squaring the current makes every instant positive — a negative current squared gives the same heat as a positive one of the same magnitude. So instead of averaging the current (which gives zero), we average the square of the current, then take the square root to get back to a current-like number. That's the root-mean-square: Root of the Mean of the Square.
For a sinusoidal current i(t)=I0sin(ωt), the square is I02sin2(ωt). The average of sin2 over a full cycle is exactly 1/2. So:
mean of i2=I02×21
Then:
Irms=2I02=2I0
Irms=2I0andVrms=2V0
The Physical Meaning
If you take a resistor and pass a sinusoidal current of peak value I0 through it, the average power dissipated is exactly the same as if you passed a steady DC current of I0/2 through it. That's why RMS is called the "equivalent DC" value.
Tip
When you see "230 V AC" on a household outlet, that 230 V is the RMS voltage. The peak voltage is 230×2≈325 V. The wire insulation has to handle 325 V peaks, but the heating effect is the same as 230 V DC.
Peak Value
The peak valueI0 (or V0) is simply the maximum instantaneous value the waveform reaches. For a sine wave, it's the amplitude. The RMS value is always smaller than the peak — by a factor of 2 for a pure sine wave. …
The house line carries sinusoidal AC, so its average current over a cycle is zero (a). The quoted 220 V is the rms value, not the average (the average voltage is also zero), which rules out (b). A 90∘ phase difference needs a purely reactive load; real household loads have resistance, so th …
The mains supply is AC, so its average current (and average voltage) over a cycle is zero, and the household load is partly resistive so the voltage–current phase difference obeys ∣ϕ∣<π/2. Correct options: (a) and (d).
Concept understanding. Domestic supply is sinusoidal AC. Over a full cycle the mean of sinωt is zero, so both the average current and the average voltage are zero.
Why (a): Average current =⟨I0sinωt⟩=0 over a cycle — true.
Why not (b): 220 V is the rms voltage, not the average; the average voltage over a cycle is zero. False.
Why not (c): A 90∘ phase difference occurs only for a purely reactive (ideal inductor or capacitor) load. Real household loads always include resistance, so the phase is not exactly 90∘. …
Method: Distinguishing Average vs RMS Quantities, and Bounding the Phase of a Realistic Load
Use this whenever a question about a general AC supply line asks you to evaluate claims mixing "average" values, "rms" values, and the voltage-current phase difference for a real (not idealized) load.
Steps
Step 1: Recall that a sinusoid's own time-average is zero over a full cycle
For any x(t)=X0sin(ωt+θ), ⟨x⟩ over one period T is exactly zero — the positive and negative halves cancel exactly. This applies to BOTH the current and the voltage of an AC supply line, independent of their amplitudes.
Step 2: Separate this from the RMS value, which is never zero
Any quoted supply rating (e.g. "220 V") is the rms value — defined via ⟨x2⟩, which squares away the sign before averaging, so it stays positive. Do not confuse a claim about the average value (always zero for pure AC) with a claim about the rms value (the quoted rating, never zero).
Step 3: Determine the allowed phase range for a REALISTIC load, not an idealized one …
Q.An alternating current having the peak value 102A is used to heat a metal wire. To produce the same heating effect, the constant current required is
(A) 102A
(B) 5A
(C) 14A
(D) 7A
(E) 10A
›Reveal solutionSolution
The equivalent steady current equals the RMS value, which is 10 A.
Concept and Intuition
Heating depends on the mean of the square of the current, so the constant current giving the same heat is the RMS value of the AC. For a sinusoid, RMS = peak/sqrt(2).
Q.An electric appliance draws 3A current from a 200 V, 50 Hz power supply. The amplitude of the supply voltage is nearly:
(A) 140 V
(B) 200 V
(C) 283 V
(D) 67 V
(E) 600 V
›Reveal solutionSolution
The stated 200V is the rms value; the peak (amplitude) is 2 times it, ≈283V.
AC supplies are specified by their rms voltage. The amplitude (peak) is
Q.An alternating current having peak value 14.14 A is used to heat a metal wire. The value of the direct current $i$ required to produce the same heating effect in the same wire is
(A) 0.707 A
(B) 28.28 A
(C) 7.07 A
(D) 10 A
(E) 14 A
Q.The INCORRECT statement is
(A) The direction of eddy currents is given by Lenz' law.
(B) A choke coil is a pure inductor used for controlling current in an A.C. circuit.
(C) The r.m.s. value of A.C. current is 2 times the peak value of A.C. current.
(D) Quality factor is a measure of sharpness of resonance in A.C. circuit.
(E) Magnetic field energy stored in an inductor of inductance L is 21LI2.
›Reveal solutionSolution
The false statement is that r.m.s. current is 2 times the peak; in fact it is the peak divided by 2.
Concept and Intuition
For a sinusoidal AC, Irms=2I0, i.e. the r.m.s. value is smaller than the peak by a factor 2, not larger. The remaining statements (Lenz's law for eddy currents, choke coil as inductor, quality factor as sharpness of resonance, inductor energy 21LI2) are all correct.
Step-by-Step Solution
Definition: Irms=I0/2≈0.707I0.
Statement (C) claims Irms=2I0, which is the inverse relationship — false. …