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

CBSEMCQ· 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.

Showing the 12 most recent of 40 on this concept.

CBSE 2026Set A1 markMCQ
$Q. The voltage of domestic ac is 220 V. What does this represent? (A) Peak value voltage (B) Mean value voltage (C) Root mean voltage (D) Root mean square voltage$
›Reveal solutionSolution

The 220 V of domestic mains is the RMS (root-mean-square) voltage.

An AC voltage varies sinusoidally, so it is specified by an effective value that produces the same heating as an equivalent DC — the root-mean-square (RMS) value. Household ratings such as "220 V" are RMS values. The peak value is larger:

$V_{peak}=\sqrt{2}\,V_{rms}=\sqrt{2}\times220\approx311\,\text{V}.$

✓Final answer
$(D) Root mean square voltage.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. The ratio of root mean square (rms) value and peak value of an alternating current is$
(a) 1 : 1
(b) 1 : 2
(c) √2 : 1
(d) 1 : √2
›Reveal solutionSolution

For a sinusoidal alternating current i = i0 sin(omega t), the rms value is i0/root2, so the ratio rms:peak is 1:root2.

RMS (root mean square) value is defined so that it produces the same heating effect as an equivalent DC. For i = i0 sin(omega t), averaging i^2 over a cycle gives <i^2> = i0^2/2, so i_rms = i0/root2 ~ 0.707 i0. Hence i_rms : i0 = 1 : root2.

✓Final answer
$(d) 1 : root 2.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. The mean value of an alternating current in a half cycle is$
(a) I0/sqrt(2)
(b) I0/2
(c) 2*I0/pi
(d) none of these
›Reveal solutionSolution

Averaging I = I0sin(omegat) over one half cycle (0 to pi/omega) gives 2*I0/pi - this is the standard 'mean/average value of AC'.

For a sinusoidal current I = I0sin(omegat), the average over a FULL cycle is zero (positive and negative halves cancel exactly). So the 'mean value' of AC is conventionally defined over just a HALF cycle, where the current keeps one sign throughout. Averaging:

I_mean = (1/T') * integral of I0sin(omegat) dt, over one half period T' = pi/omega

Carrying out this integration gives

I_mean = 2I0/pi (approx. 0.637I0)

(For comparison, the RMS value over a full cycle is I0/sqrt(2) approx 0.707*I0, a different and larger quantity - that's why (a) is wrong here.)

✓Final answer
$(c) 2*I0/pi.$
CBSE 2026Set SEM31 markMCQ
$Q. Statement I : Direct current (DC) is less dangerous than alternating current (AC). Statement II : The rms value of the alternating current (AC) is 70\cdot7% of the peak value.$
(a) Only Statement I is true.
(b) Only Statement II is true.
(c) Both Statements I and II are true.
(d) Both Statements I and II are false.
›Reveal solutionSolution

Statement I is true (AC of equal rated voltage is generally more dangerous than DC), and Statement II is true (rms value = peak/√2 = 70·7% of peak). Hence option (c).

Statement I: For the same magnitude, alternating current is generally considered more dangerous than direct current, largely because AC can cause sustained muscular contraction and its effective (rms) value acts continuously. So DC being 'less dangerous' is accepted as true.

Statement II: For a sinusoidal AC, the rms value relates to the peak value I₀ by I_rms = I₀/√2 ≈ 0·707 I₀, i.e. 70·7% of the peak — a standard NCERT/CBSE Class 12 Physics result on alternating current.

Both statements are correct.

✓Final answer
$(c) Both Statements I and II are true$
CBSE 2026Set SEM31 markMCQ
$Q. The equation of an alternating electromotive force is E = 220 sin(100πt - π/15) V, here t is in second. Its rms value and frequency are respectively$
(a) (220/√2) V, 50 Hz
(b) 220 V, 50 Hz
(c) (220/√2) V, 100 Hz
(d) 220√2 V, 50 Hz
›Reveal solutionSolution

Compare E = 220 sin(100πt − π/15) with E = E₀ sin(ωt + φ): E₀ = 220 V and ω = 100π. Then E_rms = E₀/√2 = 220/√2 and f = ω/2π = 50 Hz. Option (a).

Step 1 — peak value: E₀ = 220 V, so the rms value E_rms = E₀/√2 = 220/√2 V.

Step 2 — angular frequency: ω = 100π rad/s.

Step 3 — frequency: f = ω/(2π) = 100π/(2π) = 50 Hz.

Reading amplitude and frequency from a sinusoidal emf equation is a standard NCERT/CBSE Class 12 Physics skill (Alternating Current).

✓Final answer
$(a) 220/\sqrt2 V, 50 Hz$
CBSE 2026Set SEM31 markMCQ
$Q. The ratio of rms value and average value of current for a half cycle of an AC circuit is$
(a) √2 : π
(b) √2 : 1
(c) 2√2 : π
(d) π : 2√2
›Reveal solutionSolution

For a sinusoid, I_rms = I₀/√2 and the half-cycle average I_avg = 2I₀/π. Their ratio is (1/√2)/(2/π) = π/(2√2), i.e. π : 2√2. Option (d).

Step 1 — rms value of a sinusoidal current: I_rms = I₀/√2.

Step 2 — average value over a half cycle: I_avg = 2I₀/π (NCERT/CBSE Class 12 Physics, Alternating Current).

Step 3 — form the ratio:

I_rms/I_avg = (I₀/√2)/(2I₀/π) = (1/√2)·(π/2) = π/(2√2).

Step 4 — so I_rms : I_avg = π : 2√2 (this equals the form-factor relation, ≈ 1·11).

✓Final answer
$(d) π : 2\sqrt2$
CBSE 2025Set 55/4/11 markMCQ
$10\sqrt{2}$
›Reveal solutionSolution
An AC ammeter reads the RMS (root-mean-square) value of current. For a sinusoidal AC, the peak (maximum) current is $\sqrt{2}$ times the RMS value. Given RMS = 10 A, the maximum current is $10\sqrt{2}$ A.

The key here is understanding what an AC ammeter actually measures. Unlike a DC ammeter, which reads the average current, an AC ammeter is calibrated to read the RMS value of the current. For a sinusoidal alternating current, the RMS value is the "effective" value — it tells you the equivalent DC current that would produce the same heating effect in a resistor.

The relationship between the RMS value ( $I_{\text{rms}}$ ) and the peak or maximum value ( $I_0$ ) for a pure sine wave is:
$I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \quad \text{or} \quad I_0 = I_{\text{rms}} \times \sqrt{2}$
This comes from averaging the square of the sine function over one cycle. The factor $\sqrt{2}$ (approximately 1.414) is a fixed mathematical result for sinusoidal waveforms.

Now let's apply this directly to the problem.

The ammeter reading is given as 10 A. Since it's an AC ammeter, this is the RMS current: $I_{\text{rms}} = 10$ A.

We want the maximum instantaneous current, which is the peak value $I_0$ . Using the formula above:

$I_0 = I_{\text{rms}} \times \sqrt{2} = 10 \times \sqrt{2} \text{ A}$

That's it. No further calculation needed. The maximum value is simply $10\sqrt{2}$ amperes.

Watch out

A common mistake is to divide by $\sqrt{2}$ instead of multiplying. Remember: the peak is larger than the RMS value, so $I_0 > I_{\text{rms}}$ . Since $\sqrt{2} \approx 1.414$ , $10\sqrt{2} \approx 14.14$ A, which makes sense — the instantaneous current swings higher than the "average" reading.

Tip

If the current were not sinusoidal (e.g., a square wave or triangular wave), the factor would be different. But for standard AC mains and most textbook problems, the waveform is assumed sinusoidal unless stated otherwise.

✓Final answer
$10\sqrt{2}$
CBSE 2025Set D1 markMCQ
$Q. The peak value of an alternating current is 10 A. Its root mean square value will be (A) 5 A (B) 7.07 A (C) 10 A (D) 14.14 A$
›Reveal solutionSolution

The rms value of a sinusoidal current is the peak value divided by √2: 10/√2 ≈ 7.07 A.

For a sinusoidal alternating current, the root-mean-square value relates to the peak (amplitude) value by

$I_{rms} = \frac{I_0}{\sqrt{2}}$

With $I_0 = 10\,A$ :

$I_{rms} = \frac{10}{\sqrt{2}} = \frac{10}{1.414} \approx 7.07\,A$

✓Final answer
$(B) 7.07 A.$
CBSE 2025Set A1 mark
$Q. Write True or False: Average value of alternating current is zero for full cycle.$
›Reveal solutionSolution

The statement is True: over a full cycle the positive and negative halves of an AC waveform exactly cancel, giving zero average.

For a sinusoidal alternating current $i = I_0\sin\omega t$ , the average value over one complete cycle is found by integrating over a full period T:

$I_{avg} = \dfrac{1}{T}\int_0^T I_0\sin\omega t\, dt = 0$

This is because the current is positive for exactly one half of the cycle and equally negative for the other half, so the net area (and hence the average) over the full cycle is zero. (Note: the average value quoted in problems, $2I_0/\pi$ , refers to the average over only HALF a cycle — over a FULL cycle it is genuinely zero, which is why AC ammeters instead measure the rms value.)

✓Final answer
$True.$
CBSE 2025Set A1 mark
$Q. Write answer in one sentence: Write the root mean square value of alternating current.$
›Reveal solutionSolution

The rms value of a sinusoidal AC is I₀/√2, about 70.7% of its peak value.

For an alternating current $i = I_0\sin\omega t$ , the root-mean-square (rms) value is obtained by taking the square root of the mean of $i^2$ over one complete cycle:

$I_{rms} = \sqrt{\dfrac{1}{T}\int_0^T I_0^2\sin^2\omega t\, dt} = \dfrac{I_0}{\sqrt{2}} \approx 0.707\,I_0$

This is the effective/virtual value of AC — the direct current which would produce the same average heating (power dissipation) in a resistor as the given alternating current. It is what standard AC ammeters and voltmeters actually read.

✓Final answer
$I_rms = I₀/\sqrt2 \approx 0.707 I₀.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. If the peak value of alternating voltage in a circuit is E0, then the root mean square value will be$
(a) E0/2
(b) E0
(c) E0/sqrt(2)
(d) E0^2/2
›Reveal solutionSolution

For a sinusoidal quantity, the root-mean-square value is obtained by averaging the square of the waveform over a cycle and taking the square root, which gives peak/sqrt(2).

For e(t) = E0 sin(wt), the mean of sin^2(wt) over a full cycle is 1/2. So:

E_rms = sqrt( mean of e^2 ) = sqrt( E0^2 x 1/2 ) = E0 / sqrt(2)

This rms value is the effective DC-equivalent voltage that would deliver the same average power to a resistor - it is why household "220 V AC" refers to the rms value, not the peak.

✓Final answer
$(c) E0/sqrt(2).$
CBSE 2025Set ANNUAL1 mark
$Q. The equation of an alternating current is I = 15 sin 100t. Find its root mean square value.$
›Reveal solutionSolution

Comparing I = 15 sin(100t) with the standard form I = I0 sin(wt) gives peak current I0 = 15 A, and the rms value of any sinusoidal current is I0/sqrt(2).

The given alternating current is I = 15 sin(100t) A, matching the standard form I = I0 sin(wt) with:

I0 = 15 A (peak/amplitude), w = 100 rad/s

For a sinusoidal current, the rms (root mean square) value is related to the peak value by:

I_rms = I0 / sqrt(2)

Substituting I0 = 15 A:

I_rms = 15 / sqrt(2) = 15 / 1.414 approx 10.6 A

✓Final answer
$I_rms = 15/sqrt(2) approx 10.6 A.$

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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

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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