Q.If the effective (rms) current in a 50 cycle AC circuit is 5 A, what is the peak value of the current? What is the current 1/600 s after it was zero?
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] i0=2irms=2(5)=7.07 A. At t=1/600s, ωt=2π(50)/600=π/6, so i=7.07sin(30∘)=3.535 A. [!ANSWER] Peak current = 7.07 A; current at t=1/600 s = 3.535 A.
Given the effective (rms) current irms=5 A in a 50 Hz (50 cycle) AC circuit. The peak value is i0=2irms=1.414×5≈7.07 A.\n\nStarting from zero, the current is i=i0sinωt, with ω=2πf=2π(50)=100π rad/s. At t=1/600 s, the phase is ωt=100π×6001=6π rad (30∘). So i=7.07sin(30∘)=7.07×0.5=3.535 A. [!ANSWER] Peak current i0=7.07 A; current at t=1/600 s after zero =3.535 A.
Convert rms to peak current via i0=2irms, then evaluate i=i0sinωt at the given time using ω=2πf.
Using f directly as ω instead of first multiplying by 2π, which would give a badly wrong phase angle.
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).