Animated Solution for Physics - Electromagnetic Induction: The alternating current is given by $i=\left\{\sqrt{42}\sin \left(\frac{2\pi }{T}t\right)+10\right\}\text{A}$ The rms value of this current is ...... A.

The problem presents us with a fascinating scenario: an electrical current that isn't just a simple alternating wave, nor is it a steady direct current. Instead, it's a beautiful superposition of both!

Let's take a close look at the given equation for the current:

At first glance, this might look like a standard AC equation that got a little too complicated. But if we break it down, it reveals its secrets. The equation consists of two distinct parts added together.

The first part, $\sqrt{42}\sin \left(\frac{2\pi }{T}t\right)$, is our classic alternating current (AC). It oscillates back and forth, constantly changing direction. The second part is simply the constant $10$. This represents a direct current (DC) component.

Visually, if you were to graph this, the DC component acts as a new baseline. Instead of the sine wave oscillating around zero, the entire wave is shifted upwards by $10$ units.

To find the effective or Root Mean Square (RMS) value of this combined current, we need to analyze the components individually.

First, let's look at the AC component. Its peak amplitude is $I_0=\sqrt{42}\text{ A}$. We know that for a pure sinusoidal wave, the RMS value is the peak value divided by $\sqrt{2}$.

Next, we consider the DC component. Because a direct current is perfectly constant, its RMS value is exactly equal to its magnitude.

Here is where the magic happens. When you have independent, orthogonal components like an AC wave and a steady DC line, their powers add up linearly. Since power is proportional to the square of the RMS current, the net RMS current is the square root of the sum of the squares of the individual RMS values. It's exactly like applying the Pythagorean theorem to electrical currents!

Now, it's just a matter of plugging in the numbers we found.

Squaring the terms gives us:

And taking the square root yields our final, elegant answer:

$$I_{rms}=11\text{ A}$$

The effective RMS value of this superimposed current is $11\text{ A}$. Notice how the addition of the AC ripple slightly increases the effective heating value compared to the $10\text{ A}$ DC baseline alone. This is a powerful concept that frequently appears in advanced circuit analysis!