A DC of 5 A produces the same heating effect as an A.C. of : (a) 50 A rms current (b) 5 A peak current (c) 5 A

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Q. A DC of 5 A produces the same heating effect as an A.C. of :

(a) 50 A rms current

(b) 5 A peak current

(d) none of these

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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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By the very definition of the rms value of AC, a 5 A rms alternating current produces the same heating effect as a 5 A direct current.

The root-mean-square (rms or virtual/effective) value of an alternating current $I_{rms}$ is defined as that value of steady direct current which, when passed through a given resistor for a given time, produces the same amount of heat as is produced by the alternating current when passed through the same resistor for the same time. Mathematically, for a sinusoidal current $i = I_0\sin\omega t$ , $I_{rms} = I_0/\sqrt2$ , and this is precisely the value that satisfies $I_{rms}^2 R T = \int_0^T i^2 R\, dt$ (equal heat, hence equal average power, $P = I_{rms}^2 R$ ).

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Q.A DC of 5 A produces the same heating effect as an A.C. of :

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