Q.If the rms current in a 50 Hz AC circuit is 5 A, the value of the current 1/300 second after its value becomes zero is (A) 25 A (B) 523 A (C) 65 A (D) 52 A
Because an alternating current or voltage constantly changes both its magnitude and its sign throughout each cycle, a single steady number is needed to usefully describe its 'typical' size -- and two different such numbers are standard, each answering a different practical question. The AVERAGE value, iav=0.637i0 (equivalently eav=0.637e0), is defined over only a HALF cycle, since the average taken over a FULL cycle is always exactly zero (the positive and negative halves of the sinusoid cancel perfectly) and so is not a useful descriptor at all.
The RMS (root-mean-square, also called effective or virtual) value is defined differently, via the heating effect of the current: irms is the value of a steady DC current that would produce exactly the same amount of heat, in the same resistance, over the same time, as the actual alternating current does. Since heat production depends on i2 (always positive, regardless of the current's instantaneous direction), this quantity never averages to zero, making it a genuinely useful representative value. Working through the averaging gives irms=i0/2≈0.707i0 and, identically, erms=e0/2. Ordinary AC ammeters and voltmeters are built to display this rms value -- which is exactly why a '220 V AC' supply is understood to mean an rms value of 220 V, even though the instantaneous voltage actually swings all the way up to a peak of 2×220≈311 V within every cycle. A moving-coil meter, by contrast, responds to the (always-zero) average value and so cannot be used to measure AC directly at all.
[!TLDR] With irms=5 A, the peak current is i0=52 A. At t=1/300 s with f=50 Hz, ωt=2π(50)(1/300)=π/3, so i=i0sin(π/3)=52×23=53/2 A. [!ANSWER] (B) 53/2 A
The rms current is given as irms=5 A, so the peak current is i0=2irms=52 A. Taking the current to start from zero at t=0 and rise as i=i0sinωt, with ω=2πf=2π(50)=100π rad/s, the phase at t=1/300 s is ωt=100π×3001=3π rad (60∘). So i=i0sin(3π)=52×23=256 A. Simplifying, 256=546=523≈6.12 A. [!ANSWER] (B) 53/2 A ≈6.12 A
Recover the peak current from the given rms value (i0=2irms), then evaluate i=i0sinωt at the given instant using ω=2πf.
Substituting the rms value directly into i=irmssinωt instead of first converting to the peak value i0=2irms, which understates the answer by a factor of 2.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2023Set ANNUAL1 mark
Q.What is the average value of alternating current over a complete cycle?
›Reveal solutionSolution
Positive and negative half-cycles of a sinusoidal AC exactly cancel over one full cycle.
The mean value of i=I0sin(ωt) over one complete cycle (T1∫0Tidt) is zero, because the positive half-cycle's area is exactly cancelled by the equal-and-opposite negative half-cycle's area. This is precisely why the r.m.s. value (based on the mean of i2, which is always positive) is used to characterise the effective magnitude of an AC instead of the average value.
✓Final answer
Average value over a complete cycle is zero.
CBSE 2022Set ANNUAL1 markMCQ
Q.The average value of alternating current over a full cycle is always _____. [I0 = Peak value of current]
(a) zero
(b) I0/2
(c) I0/2
(d) 2I0
›Reveal solutionSolution
Over a full cycle, the positive and negative half-cycles of a sine wave cancel exactly.
For a sinusoidal AC, i=I0sinωt. Averaged over one complete cycle, ⟨sinωt⟩=0 because the positive half-cycle exactly cancels the negative half-cycle. Hence the average value of AC over a full cycle is always zero (the average over a half-cycle, 2I0/π, is the physically meaningful non-zero average).