Question 10: AC fundamentals

Question

A direct current of $$10 \mathrm{~A}$$ is superimposed on an alternating current $$i=10 \sin \omega t$$ flowing through the wire. The effective value of the resulting current will be.

Accepted Answer

The resulting current is the sum of the direct current and the alternating current:
$$I = I_{dc} + I_{ac} = 10 + 10 \sin \omega t$$
To find the effective value (also known as the root-mean-square or RMS value), we need to calculate the square root of the mean of the squared current:
$$I_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} I^2 dt}$$
where T is the period of the alternating current.
Let's substitute the expression for I and simplify:
$$I_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} (10 + 10 \sin \omega t)^2 dt}$$
Expanding the square and integrating term by term:
$$I_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} (100 + 200 \sin \omega t + 100 \sin^2 \omega t) dt}$$
We know that the average value of  $$ \sin \omega t $$ over a complete cycle is zero, and the average value of  $$ \sin^2 \omega t $$ over a complete cycle is  $$\frac{1}{2}$$. Therefore:
$$I_{rms} = \sqrt{\frac{1}{T} (100T + 0 + 100 \cdot \frac{1}{2} T)} = \sqrt{150}$$
Simplifying further:
$$I_{rms} = 5 \sqrt{6} \mathrm{~A}$$
Therefore, the correct answer is
Option B:  $$5 \sqrt{6} \mathrm{~A}$$
.

Answer

$$(15 / 2) \mathrm{A}$$

Answer

$$5 \sqrt{6} \mathrm{~A}$$

Answer

$$5 \sqrt{5} \mathrm{~A}$$

Answer

$$15 \mathrm{~A}$$

AC fundamentals

A direct current of $10 \mathrm{~A} is superimposed on an alternating current i=10 \sin \omega t$ flowing through the wire. The effective value of the resulting current will be.

Choose the correct answer: