If the rms current in a 50 Hz ac circuit is 5 A, the value of the current $\dfrac{1}{300}$ seconds after its value becomes zero is (a) 5√2 A (b) 5√3/2 A (c) 5/6 A (d) 5/√2 A

If the rms current in a 50 Hz ac circuit is 5 A, the value of the current (1)/(300) seconds after its value be

NCERT Exemplar · Q1

\dfrac{1}{300}

(a) 5√2 A

(b) 5√3/2 A

(d) 5/√2 A

Andhra Pradesh BieapMCQ· 1mImportance★★★★★

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✓ Free question

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.

Watch out

The $\sqrt{2}$ factor applies only to sinusoidal waveforms. For a square wave, $I_{\text{rms}} = I_0$ ; for a triangular wave, $I_{\text{rms}} = I_0/\sqrt{3}$ . Never blindly use $/\sqrt{2}$ unless you know the waveform is sinusoidal.

Summary

Quantity	Symbol	Meaning
Peak current	$I_0$	Maximum instantaneous current
RMS current	$I_{\text{rms}} = I_0/\sqrt{2}$	Equivalent DC that gives same heating
Peak voltage	$V_0$	Maximum instantaneous voltage
RMS voltage	$V_{\text{rms}} = V_0/\sqrt{2}$	Equivalent DC voltage for same power

The core idea: RMS converts an alternating quantity into a steady DC equivalent for power calculations. It's the square root of the average of the square — nothing more, nothing less.

RMS and peak value calculations open the NCERT Class 12 Physics chapter on Alternating Current, and 'RMS value formula class 12 physics' or 'AC RMS and peak value important questions' are frequently searched by board and JEE Main aspirants. Because household AC ratings are always quoted as RMS values, this concept also shows up in applied, real-world exam questions.

Convert rms to peak, then evaluate the sine at the given instant. The current $\tfrac{1}{300}$ s after a zero-crossing is $5\sqrt{\tfrac{3}{2}}\ \text{A}\approx 6.12\ \text{A}$ .

For a sinusoidal current, the peak value is

I_0 = \sqrt{2}\,I_{\text{rms}} = 5\sqrt{2}\ \text{A}.

Take the instant of zero as $t=0$ , so $i(t)=I_0\sin(\omega t)$ with $\omega = 2\pi f = 2\pi(50)=100\pi\ \text{rad/s}$ .

At $t=\dfrac{1}{300}\ \text{s}$ :

i = 5\sqrt{2}\,\sin\!\left(100\pi\cdot\frac{1}{300}\right)=5\sqrt{2}\,\sin\!\left(\frac{\pi}{3}\right)=5\sqrt{2}\cdot\frac{\sqrt{3}}{2}=\frac{5\sqrt{6}}{2}=5\sqrt{\frac{3}{2}}\ \text{A}.

Numerically, $i \approx 6.12\ \text{A}$ . (Note that $\tfrac{1}{300}\,\text{s}=\tfrac{T}{6}$ of the period $T=0.02\,\text{s}$ , i.e. a phase of $60^\circ$ .)

✓Final answer

5\sqrt{\dfrac{3}{2}}\ \text{A}\approx 6.12\ \text{A}

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

$\sin\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2}$ .

$V(t) = V_0\sin(\omega t) = \dfrac{200}{\sqrt2}\times\dfrac12 = \dfrac{100}{\sqrt2}$ V.

Common Mistakes

Mistaking the given $V_0=\dfrac{200}{\sqrt2}$ for an rms value and dividing by $\sqrt2$ again (that value is already the peak, as stated).

Working in degrees instead of radians when evaluating $\sin(\omega t)$ , or mis-simplifying $100\pi/600$ .

✓Final answer
$\dfrac{100}{\sqrt{2}}\ V$

ANSWER: C
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).

$t' = \dfrac{\pi/4}{\omega} = \dfrac{\pi/4}{10\pi} = \dfrac{1}{40}\,s = 2.5\times 10^{-2}\,s$ .

Common Mistakes

Using $\sin$ instead of $\cos$ when measuring time from the peak (a sine measured from t=0 reaches its peak only at $t=T/4$ , so re-anchoring the origin to the peak is essential).

Forgetting R cancels out — R is a distractor here since we only need the phase, not the current magnitude.

✓Final answer
$2.5\times10^{-2}$

ANSWER: D
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

Adding the individual rms values of the sine and cosine parts directly (i.e., $3/\sqrt2 + 4/\sqrt2$ ) instead of first combining amplitudes — this is wrong because rms doesn't add linearly for non-identical-phase components in that naive way; the correct approach is to combine amplitudes via the Pythagorean sum first.

Forgetting the $1/\sqrt2$ factor altogether and answering with the peak value.

✓Final answer
$\dfrac{5}{\sqrt2}$

ANSWER: C
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.

This is closest to "nearly 1 A" among the options.

Common Mistakes

Forgetting to apply the $\sqrt2$ factor and treating the RMS current as the peak current directly.

Using $R=280\,\Omega$ with the peak voltage instead of RMS voltage when the given 200 V is explicitly the (implied) RMS supply value.

✓Final answer
$The correct option is (A) — Nearly 1 A.$

ANSWER: A

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Q.If the rms current in a 50 Hz ac circuit is 5 A, the value of the current $\dfrac{1}{300}$ seconds after its value becomes zero is

Concept understanding — RMS and Peak Value

Why We Need a New Measure

The Intuition: Squaring Fixes the Sign Problem

The Physical Meaning

Peak Value

Summary

Method: Finding the Instantaneous Value of an AC Quantity at a Given Time

Steps

Concept and Intuition

Step-by-Step Solution

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Common Mistakes

Q.If the rms current in a 50 Hz ac circuit is 5 A, the value of the current 1300\dfrac{1}{300}3001​ seconds after its value becomes zero is

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Q.If the rms current in a 50 Hz ac circuit is 5 A, the value of the current $\dfrac{1}{300}$ seconds after its value becomes zero is

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