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

Odisha ChseMCQ· 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}

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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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