I = i1 sin(ωt) + i2 cos(ωt). Find the rms value of the current.
Understand the Problem
The question is asking to find the root mean square (rms) value of the current described by the equation I = i1sin(ωt) + i2cos(ωt). This involves calculating the rms value based on the contributions of both sine and cosine components of the current.
Answer
The RMS value of the current is given by $I_{\text{rms}} = \sqrt{\frac{i_1^2 + i_2^2}{2}}$.
Answer for screen readers
The root mean square (RMS) value of the current is given by: $$ I_{\text{rms}} = \sqrt{\frac{i_1^2 + i_2^2}{2}} $$
Steps to Solve
- Identify the Current Components
The current $I$ is given as a combination of sine and cosine functions: $$ I = i_1 \sin(\omega t) + i_2 \cos(\omega t) $$
- Square the Current Expression
To find the root mean square (RMS) value, we first square the current: $$ I^2 = (i_1 \sin(\omega t) + i_2 \cos(\omega t))^2 $$
- Expand the Squared Expression
Expanding the above square using the formula $(a + b)^2 = a^2 + 2ab + b^2$ gives: $$ I^2 = i_1^2 \sin^2(\omega t) + 2i_1 i_2 \sin(\omega t) \cos(\omega t) + i_2^2 \cos^2(\omega t) $$
- Calculate the Average Value
To find the RMS value, we need to find the average value of $I^2$ over one period $T$: $$ \text{Average}(I^2) = \frac{1}{T} \int_0^T I^2 , dt $$
- Integrate Each Component
Using known integrals:
- The integral of $\sin^2(\omega t)$ over one period is $\frac{T}{2}$.
- The integral of $\cos^2(\omega t)$ over one period is also $\frac{T}{2}$.
- The integral of $\sin(\omega t) \cos(\omega t)$ over one period is $0$.
Thus: $$ \text{Average}(I^2) = \frac{1}{T} \left( i_1^2 \frac{T}{2} + i_2^2 \frac{T}{2} \right) = \frac{i_1^2 + i_2^2}{2} $$
- Calculate RMS Value
The RMS value is obtained by taking the square root of the average of $I^2$: $$ I_{\text{rms}} = \sqrt{\text{Average}(I^2)} = \sqrt{\frac{i_1^2 + i_2^2}{2}} $$
The root mean square (RMS) value of the current is given by: $$ I_{\text{rms}} = \sqrt{\frac{i_1^2 + i_2^2}{2}} $$
More Information
The root mean square (RMS) value is particularly useful in electrical engineering as it gives a measure of the effective value of an alternating current (AC). It allows comparison with direct current (DC) circuits, where the current is constant.
Tips
- A common mistake is forgetting to correctly integrate the product of sine and cosine functions, as their average over a complete cycle is zero.
- Another mistake is not properly applying the average over one period for the square of the current.
