Q.Calculate the instantaneous value at $60^{\circ}$, average value and RMS value of an alternating current whose peak value is 20 A.

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:

Then:

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.

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. …