An alternating current $I$ (in amperes) varies with time $t$ as a periodic square (piecewise constant) waveform of period $T$: during the first half of each period ($0$ to $T/2$) the current is constant at $+1$ A, and during the second half ($T/2$ to $T$) it is constant at $-2$ A, and this pattern repeats every period. Find the rms value of this current.

An alternating current I (in amperes) varies with time t as a periodic square (piecewise constant) waveform of

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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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The rms current is the square root of the time-average of $I^2$ over one period. With $I=+1$ A for the first half-period and $I=-2$ A for the second, the mean of $I^2$ is $(1+4)/2=2.5\,\text{A}^2$ , giving $I_{rms}=\sqrt{2.5}\approx1.58$ A.

Concept: rms of a periodic current

The rms (effective) value is defined by

I_{rms}=\sqrt{\langle I^2\rangle}=\sqrt{\frac{1}{T}\int_0^T I^2\,dt}.

For a piecewise-constant waveform the integral is just a weighted average of the squared levels.

Steps

The waveform over one period: $I=+1$ A for $0\le t<T/2$ , and $I=-2$ A for $T/2\le t<T$ .
Square each level (squaring removes the sign): $I^2=1\,\text{A}^2$ for the first half, $I^2=4\,\text{A}^2$ for the second half.
Time-average of $I^2$ over the period: …

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Concept: rms of a periodic current

Steps

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