The devices consume a real power of

$$P = 500\,\text{W}.$$

The AC voltage is given by

$$E(t) = 200\sin\left(3.14t + \frac{\pi}{4}\right).$$

Here, the peak voltage is

$$V_m = 200\,\text{V},$$

so its rms value is

$$V_{rms} = \frac{V_m}{\sqrt{2}} = \frac{200}{\sqrt{2}}\ \text{V}.$$

In AC circuits the real power is given by the formula

$$P = \frac{1}{2}V_m I_m \cos\phi,$$

where

$$I_m$$ is the peak current, and

$$\phi$$ is the phase difference between the voltage and the current.

In this problem the voltage is written with a phase angle of $$\pi/4$$. This phase angle is usually associated with the load’s impedance and hence the phase difference between voltage and current. Assuming that the current has zero phase (i.e. taken as the reference), then

$$\phi = \frac{\pi}{4}.$$

And note that

$$\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}.$$

Substitute the given values into the power equation:

$$500 = \frac{1}{2} \times 200 \times I_m \times \frac{1}{\sqrt{2}}.$$

Simplify this step-by-step:

First, compute the constant:

$$\frac{1}{2} \times 200 = 100.$$

Then the equation becomes:

$$500 = 100 \times I_m \times \frac{1}{\sqrt{2}}.$$

Solve for $$I_m$$:

$$I_m = \frac{500\sqrt{2}}{100} = 5\sqrt{2}\ \text{A}.$$

Now, convert the peak current to rms current using

$$I_{rms} = \frac{I_m}{\sqrt{2}}.$$

Substitute $$I_m= 5\sqrt{2}\ \text{A}:$$

$$I_{rms} = \frac{5\sqrt{2}}{\sqrt{2}} = 5\ \text{A}.$$

Thus, the rms value of the circuit current is 5 A, which corresponds to Option D.

In summary:

Given the voltage, the power is calculated using the peak values and the cosine of the phase difference.

After finding the peak current, it is converted to its rms value.

The final answer is $$\boxed{5\,\text{A}}.$$

Alternating Current

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