Calculate the active power

Question

I'm asked to determine the active power (P). (I don't know if that is actually called active power in English?)

The formula of P is = U * I * sin(alpha)

How can we determine the alpha in this case, so that we can calculate the P?

Answer 1

Well, we have the following circuit:

^{simulate this circuit – Schematic created using CircuitLab}

The input voltage is given in the following diagram:

Assuming an ideal diode, the negative parts are cut of for the voltage across the resistor. That is shown in the following diagram:

Now, we know that the power in a resistor is given by:

$$\text{P}_\text{R}\left(t\right)=\frac{\text{V}_\text{R}^2\left(t\right)}{\text{R}}\tag1$$

It is not hard to show that \$\text{V}_\text{R}\left(t\right)\$, is given by:

$$\text{V}_\text{R}\left(t\right)=\frac{\text{V}}{\text{T}}\sum_{\text{n}\ge0}\theta\left(t-\text{Tn}\right)\mathcal{I}_\text{n}\left(t,\text{T}\right)\tag2$$

Where:

$$\mathcal{I}_\text{n}\left(t,\text{T}\right)=\left(\text{T}\left(1+2\text{n}\right)-2t\right)\left(\theta\left(t-\text{T}\left(\frac{1}{2}+\text{n}\right)\right)+\theta\left(t-\text{T}\left(1+\text{n}\right)\right)\right)\tag3$$

Now, we can look at the average power and the RMS power using the following two formulas:

• Average power: $$\overline{\text{P}}_{\text{R}}=\frac{1}{\text{T}}\int_0^\text{T}\text{P}_\text{R}\left(t\right)\space\text{d}t\tag4$$
• RMS-power: $$\text{P}_{\text{R}|\text{RMS}}=\sqrt{\frac{1}{\text{T}}\int_0^\text{T}\text{P}_\text{R}^2\left(t\right)\space\text{d}t}\tag5$$

I used Mathematica to find them:

Average power:

In[1]:=Integrate[(1/
    T)*(((Sum[(V/
           T)*((HeavisideTheta[
             t - T*n])*(((-2 t + T + 
               2 n T) (-HeavisideTheta[t - T/2 - n T] + 
               HeavisideTheta[t - (1 + n) T]))/1)), {n, 0, 
         Infinity}])^2)/R), {t, 0, T}, Assumptions -> T > 0]

Out[1]=V^2/(6 R)

RMS-power:

In[2]:=FullSimplify[
 Sqrt[(1/T)*
   Integrate[((((Sum[(V/
               T)*((HeavisideTheta[
                 t - T*n])*(((-2 t + T + 
                   2 n T) (-HeavisideTheta[t - T/2 - n T] + 
                   HeavisideTheta[t - (1 + n) T]))/1)), {n, 0, 
             Infinity}])^2)/R))^2, {t, 0, T}, Assumptions -> T > 0]], 
 Assumptions -> V > 0 && R > 0]

Out[2]=V^2/(Sqrt[10] R)

Answer 2

You'll notice that your circuit is nonlinear, due to the diode. Hence, if you want to know powers, you'll have to integrate the squared product of current and voltage over a period and then divide the thus calculated energy by the length of the period.

You seem to have a description of the voltage over time (although that graph isn't sufficiently labeled, so we don't know what we're looking at exactly; this might be the voltage over the diode, over the resistor, or over the voltage source), but you don't have a description of the current. You need to first find that definition of the current over time!

That might be very easy, or relatively complex, depending on how you model that diode. But it's something we can't do for you.

The rest should just be solving an integral. That's doable.

Answer 3

I'm asked to determine the active power (P)

And

The formula of P is = U * I * sin(alpha)

---

Not in this universe. That formula is for "so-called" reactive power but, you don't have any reactive components so it's null and void.

---

The RMS of a triangle or saw-tooth waveform is this: -

So, without the diode half wave rectifier, the power is: -

$$\dfrac{\left[\dfrac{V_P}{\sqrt3}\right]^2}{R}$$

But, because of the rectifier, only half of that power reaches the load: -

$$\dfrac{\left[\dfrac{V_P}{\sqrt3}\right]^2}{2\cdot R}$$

I'll leave you to plug in the numbers.