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

Karnataka PUCMCQ· 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.

COMEDK 2023Set 2023-E1 markMCQ
$220 \mathrm{~V}$
›Reveal solutionSolution

(The other options are not physical: shock from a.c. is not 'always repulsive', d.c. has zero frequency but that is not the reason, and 'speed of ac/dc' is meaningless here.)

Concept: an a.c. supply is quoted by its RMS value, but the instantaneous voltage swings up to the PEAK value.

For 220 V a.c. (rms):

V_peak = sqrt2 x V_rms = 1.414 x 220 = 311 V.

For 220 V d.c. the voltage is steady at 220 V.

So the a.c. mains repeatedly reaches about 311 V - roughly 1.4 times the d.c. value - and it is this higher peak that drives a larger peak current through the body, making 220 V a.c. more dangerous.

(The other options are not physical: shock from a.c. is not 'always repulsive', d.c. has zero frequency but that is not the reason, and 'speed of ac/dc' is meaningless here.)

✓Final answer
$The correct option is (A) — The peak value of ac is greater than the given value of dc$

ANSWER: A
COMEDK 2022Set 20221 markMCQ
$Q. A DC ammeter and a hot wire ammeter are connected to a circuit in series. When a direct current is passed through circuit, the DC ammeter shows 6 A. When AC current flows through circuit, what is the average readings in DC ammeter and the AC ammeter, if DC and AC currents flows simultaneously through the circuit? (A) DC = 6 A, AC = 10 A (B) DC = 3 A, AC = 5 A (C) DC = 5 A, AC = 8 A (D) DC = 2 A, AC = 3 A$
›Reveal solutionSolution

So DC = 6 A, AC (hot-wire) = 10 A.

Concept: A moving-coil DC ammeter responds to the average (mean) current; a hot-wire ammeter responds to the rms current (it is heat-operated, ∝ i²).

With DC alone the DC meter reads 6 A → I_dc = 6 A. With AC alone the hot-wire meter reads the rms value of the AC — from the option set this is 8 A.

Now pass both together: i(t) = I_dc + i_ac(t).

DC ammeter (average): ⟨i⟩ = I_dc + ⟨i_ac⟩ = 6 + 0 = 6 A (an AC current averages to zero).

Hot-wire ammeter (rms):

⟨i²⟩ = ⟨(I_dc + i_ac)²⟩ = I_dc² + 2I_dc⟨i_ac⟩ + ⟨i_ac²⟩ = 6² + 0 + 8² = 36 + 64 = 100

I_rms = √100 = 10 A.

So DC = 6 A, AC (hot-wire) = 10 A.

✓Final answer
$The correct option is (A) — DC = 6 A, AC = 10 A$

ANSWER: A
COMEDK 2022Set 20221 markMCQ
$Q. The AC voltage across a resistance can be measured using a (A) hot wire voltmeter (B) moving coil galvanometer (C) potential coil galvanometer (D) moving magnetic galvanometer$
›Reveal solutionSolution
[!TLDR]

AC needs an instrument whose reading does not depend on current direction; the hot-wire voltmeter uses the heating effect and measures the RMS AC voltage.

Concept

This CBSE Class 12 Alternating Current idea distinguishes instruments by what they respond to. A moving-coil galvanometer measures the mean value of current; since the mean of a symmetric AC is zero, it reads zero for AC. A hot-wire instrument depends on $I^2R$ heating, which is always positive regardless of direction, so it works for both AC and DC.

Solution

Moving-coil (and moving-magnet) galvanometers give a deflection proportional to the average current, which vanishes over an AC cycle, so options (B) and (D) cannot measure AC.

Option (C) 'potential coil galvanometer' is not a standard AC-measuring instrument for this purpose.

A hot-wire voltmeter passes the current through a thin wire; the wire heats up by $I^2R$ and expands, moving the pointer. Because heating is independent of the current's direction, it registers AC (its RMS value) as well as DC.

[!ANSWER]

(A) hot wire voltmeter

Note

This solution was worked out by our team and independently cross-checked by a second solve. The official answer key on record for this question could not be confirmed, so please cross-verify with the official paper where possible.
COMEDK 2021Set 2021-B1 markMCQ
$Q. A 20 ohm electric heater is connected to a 220 V, 60 Hz mains supply. The peak value of electric current flowing in the circuit is (A) 22 A (B) 5.5 A (C) 11 A (D) 15.55 A$
›Reveal solutionSolution

$I_{rms}=220/20=11$ A, peak $=\sqrt2\times11\approx15.55$ A.

For a purely resistive heater, the rms current is
$I_{rms} = \frac{V_{rms}}{R} = \frac{220}{20} = 11\ \text{A}.$
The peak (maximum) value of an AC current is $\sqrt2$ times the rms value:
$I_0 = \sqrt2\,I_{rms} = 1.414 \times 11 \approx 15.55\ \text{A}.$
✓Final answer
$The correct option is (D) — 15.55 A$
COMEDK 2021Set 2021-B1 markMCQ
$V = 200 \sin 314t$
›Reveal solutionSolution
$f = 314/2\pi = 50$ Hz; $V_{rms} = 200/\sqrt2 = 141$ V.

Comparing $V = 200\sin(314t)$ with $V = V_0\sin(\omega t)$ :

Angular frequency $\omega = 314$ rad/s, so

$f = \frac{\omega}{2\pi} = \frac{314}{6.283} \approx 50\ \text{Hz}.$

Peak voltage $V_0 = 200$ V, so

$V_{rms} = \frac{V_0}{\sqrt2} = \frac{200}{1.414} \approx 141\ \text{V}.$
✓Final answer
$The correct option is (D) — 50 Hz, 141 V$

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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

Solution

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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