prove that I=square root of (I^2avarage). using I=Iosin(wt)
To prove the equation I = √(I^2_avg) using I = Io*sin(wt), we need to work through a few steps.
Step 1: Calculate the average value of I^2 over one complete cycle.
- Substitute I = Io*sin(wt) into I^2 to get I^2 = Io^2 * sin^2(wt).
- Integrate I^2 over one complete cycle using the limits 0 to 2π to get:
I^2_avg = (1/2π) * ∫[0-2π] Io^2 * sin^2(wt) dt.
Step 2: Simplify the integral.
- Use the trigonometric identity sin^2(x) = (1/2) * (1 - cos(2x)) to simplify I^2_avg.
I^2_avg = (1/2π) * ∫[0-2π] Io^2 * [1 - cos(2wt)] dt.
- Distribute Io^2 and split the integral into two parts:
I^2_avg = (Io^2/2π) * ∫[0-2π] [1 - cos(2wt)] dt.
Step 3: Perform the integration.
- Integrate the first term, 1, with respect to t over the limits 0 to 2π.
∫[0-2π] 1 dt = [t] [0-2π] = 2π - 0 = 2π.
- Integrate the second term, -cos(2wt), with respect to t over the limits 0 to 2π.
∫[0-2π] -cos(2wt) dt = [-(1/2w) * sin(2wt)] [0-2π] = 0 - (-(1/2w) * sin(4πw)) = 0.
Step 4: Simplify the equation.
- Substitute the integrated values back into the equation for I^2_avg:
I^2_avg = (Io^2/2π) * (2π - 0) = Io^2.
Step 5: Take the square root of I^2_avg.
- Taking the square root of both sides gives:
I = √(I^2_avg) = √(Io^2) = Io.
Therefore, we have proved that I = √(I^2_avg) using I = Io * sin(wt) with the average value of I^2 over one complete cycle.