The line that draws power supply to your house from street has (a) zero average current. (b) 220 V average voltage. (c) voltage and current out of phase by 90°. (d) voltage and current possibly differing in phase φ such that |φ| < π/2.

The line that draws power supply to your house from street has (a) zero average current. (b) 220 V average vol

NCERT Exemplar · Q31

Q. The line that draws power supply to your house from street has

(a) zero average current.

(b) 220 V average voltage.

(d) voltage and current possibly differing in phase φ such that |φ| < π/2.

Andhra Pradesh BieapMCQ· 1mImportance★★★★★

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Concept understanding — RMS and Peak Value

Why We Need a New Measure

When you push a DC current through a resistor, the power is constant — $P = I^2 R$ , and the heating is steady. But an AC current keeps changing direction and magnitude. At one instant it's $+I_0$ , a moment later it's zero, then $-I_0$ . 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 $I^2$ , 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 $i(t) = I_0 \sin(\omega t)$ , the square is $I_0^2 \sin^2(\omega t)$ . The average of $\sin^2$ over a full cycle is exactly $1/2$ . So:

\text{mean of } i^2 = I_0^2 \times \frac{1}{2}

Then:

I_{\text{rms}} = \sqrt{\frac{I_0^2}{2}} = \frac{I_0}{\sqrt{2}}

I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \quad \text{and} \quad V_{\text{rms}} = \frac{V_0}{\sqrt{2}}

The Physical Meaning

If you take a resistor and pass a sinusoidal current of peak value $I_0$ through it, the average power dissipated is exactly the same as if you passed a steady DC current of $I_0/\sqrt{2}$ 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 $230 \times \sqrt{2} \approx 325$ 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 value $I_0$ (or $V_0$ ) 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 $\sqrt{2}$ for a pure sine wave. …

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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 $|\phi|<\pi/2$ . Correct options: (a) and (d).

Concept understanding. Domestic supply is sinusoidal AC. Over a full cycle the mean of $\sin\omega t$ is zero, so both the average current and the average voltage are zero.

Why (a): Average current $=\langle I_0\sin\omega t\rangle = 0$ 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 $90^\circ$ 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 $90^\circ$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
$\dfrac{200}{\sqrt{2}}$
›Reveal solutionSolution
Plug $t=1/600$ s into $V(t)=V_0\sin(\omega t)$ — the phase works out to $\pi/6$ , halving the peak voltage, giving $\dfrac{100}{\sqrt2}$ V.

Concept and Intuition

An AC voltage varies as $V(t) = V_0\sin(\omega t)$ , where $V_0$ is the peak (amplitude), not the rms value. Here $V_0=\dfrac{200}{\sqrt2}$ V is explicitly stated to be the peak voltage — that phrase tells us not to divide by $\sqrt2$ again. The only work is finding what fraction of the peak is reached at the given instant, via the angular frequency $\omega = 2\pi f$ .

Step-by-Step Solution

Angular frequency: $\omega = 2\pi f = 2\pi(50) = 100\pi\ \text{rad/s}$ .

Phase at $t=\dfrac{1}{600}$ s: $\omega t = 100\pi\times\dfrac{1}{600} = \dfrac{100\pi}{600} = \dfrac{\pi}{6}$ .

$\sin\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2}$ . …
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AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
$25 \times 10^{-1}$
›Reveal solutionSolution
Measuring time from the current's peak, the current falls to its rms value $(I_0/\sqrt2)$ after

a phase of $\pi/4$ , which corresponds to $t=2.5\times10^{-2}\,s$ for $\omega=10\pi$ .

Concept and Intuition

For a sinusoidal current $i(t)=I_0\sin(\omega t)$ , the rms value is $I_0/\sqrt2$ . Starting the

clock at the instant the current is at its peak, the current follows $i=I_0\cos(\omega t')$

(cosine, since it's momentarily flat at the peak). We need the phase at which this cosine equals

$1/\sqrt2$ .

Step-by-Step Solution

$i(t) = \dfrac{V(t)}{R} = \dfrac{200\sin(10\pi t)}{20} = 10\sin(10\pi t)$ , so $I_0=10\,A$ , $\omega=10\pi\,\text{rad/s}$ .

Redefine time from the peak instant: $i = I_0\cos(\omega t')$ .

We want $i = I_0/\sqrt2 \Rightarrow \cos(\omega t') = 1/\sqrt2 \Rightarrow \omega t' = \pi/4$ (the first, smallest such angle after the peak). …
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AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
$i = (3\sin\omega t + 4\cos\omega t)$
›Reveal solutionSolution
Combining the sine and cosine terms into a single sinusoid of amplitude 5 A, the rms current is $5/\sqrt2$ A.

Concept and Intuition

A sum of a sine and a cosine of the same angular frequency is itself a pure sinusoid (just phase-shifted), with amplitude $\sqrt{a^2+b^2}$ for $a\sin\omega t + b\cos\omega t$ . This is because $a\sin\theta+b\cos\theta = R\sin(\theta+\phi)$ with $R=\sqrt{a^2+b^2}$ . Once we have a single sinusoid, the standard rms-to-peak relation $i_{rms}=i_0/\sqrt2$ applies directly.

Step-by-Step Solution

Write $i = 3\sin\omega t + 4\cos\omega t$ .

This is equivalent to $i = 5\sin(\omega t+\phi)$ where $5=\sqrt{3^2+4^2}$ (a 3-4-5 right triangle) and $\phi=\tan^{-1}(4/3)$ .

So the peak current is $i_0 = 5$ A.

RMS value of a sinusoidal current: $i_{rms} = \dfrac{i_0}{\sqrt2} = \dfrac{5}{\sqrt2}$ A.

Common Mistakes …
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AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
$\Omega$
›Reveal solutionSolution
Converting the given RMS AC voltage to RMS current via Ohm's law, then multiplying by $\sqrt2$ , gives a peak current of about 1 A.

Concept and Intuition

AC voltmeters and the "200 V" rating of a household-type supply refer to the RMS (root-mean-square) value, not the peak value. For a purely resistive load (a bulb), Ohm's law applies instantaneously (and hence also to RMS values), and the relationship between peak and RMS is $I_0=\sqrt2\,I_{rms}$ for a sinusoidal waveform.

Step-by-Step Solution

RMS voltage: $V_{rms}=200$ V. Resistance: $R=280\,\Omega$ .

RMS current: $I_{rms}=\dfrac{V_{rms}}{R}=\dfrac{200}{280}=0.714$ A.

Peak current: $I_0=\sqrt2\times I_{rms}=1.414\times0.714\approx1.01$ A. …
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Q.The line that draws power supply to your house from street has

Concept understanding — RMS and Peak Value

Why We Need a New Measure

The Intuition: Squaring Fixes the Sign Problem

The Physical Meaning

Peak Value

Method: Distinguishing Average vs RMS Quantities, and Bounding the Phase of a Realistic Load

Steps

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Common Mistakes …

Concept and Intuition

Step-by-Step Solution

Q.The line that draws power supply to your house from street has

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