# Finding the average power dissipated in a resistor

Well, I've a series RL circuit that is powered by a voltage source. The input voltage is given by:

$$\text{V}_{\text{in}}\left(t\right)=50+\frac{400}{\pi}\cdot\sum_{\text{n}=1}^\infty\frac{\sin\left(200\pi\cdot\text{n}\cdot t\right)}{\text{n}}\tag1$$

The resistor value is equal to $$\\text{R}=20\space\Omega\$$ and the inductor value is equal to $$\\text{L} =25\cdot10^{-3}\space\text{H}\$$.

Question: I've to find the average power that is dissipated in the resistor.

My work:

We can write for the power dissipated in the resistor, using Ohm's law, that:

$$\text{P}_\text{R}\left(t\right)=\text{V}_\text{R}\left(t\right)\cdot\text{I}_\text{R}\left(t\right)=\text{I}_\text{R}^2\left(t\right)\cdot\text{R}\tag2$$

Where $$\\text{I}_\text{R}\left(t\right)\$$ is the current trough the resistor and $$\\text{V}_\text{R}\left(t\right)\$$ is the voltage across the resistor.

Because it is a series circuit, the input current, $$\\text{I}_\text{in}\left(t\right)\$$, delivered by the source is the same trough the resistor and the inductor, so $$\\text{I}_\text{R}\left(t\right)=\text{I}_\text{in}\left(t\right)=\text{I}_\text{L}\left(t\right)\$$. Using Faraday's law, we can find that input current:

$$-\text{V}_\text{in}\left(t\right)+\text{I}_\text{in}\left(t\right)\cdot\text{R}=-\text{I}_\text{in}'\left(t\right)\cdot\text{L}\tag3$$

The initial condition is equal to $$\0\$$, so we know that $$\\text{I}_\text{in}\left(0\right)=0\$$. Now we need to solve equation $$\(3)\$$ using the values that are given.

Solving equation $$\(3)\$$ gives:

$$\text{I}_\text{in}\left(t\right)=\frac{1}{\text{L}}\cdot\exp\left(-\frac{\text{R}}{\text{L}}\cdot t\right)\cdot\left\{\int_1^t\text{X}\left(\tau\right)\space\text{d}\tau-\int_1^0\text{X}\left(\tau\right)\space\text{d}\tau\right\}\tag4$$

Where $$\\text{X}\left(\tau\right)=\exp\left(\frac{\text{R}}{\text{L}}\cdot\tau\right)\cdot\text{V}_\text{in}\left(\tau\right)\$$.

The average power that is dissipated in the resistor is equal to:

$$\overline{\text{P}}_\text{R}=\lim_{\text{n}\to\infty}\frac{1}{\text{n}}\cdot\int_0^\text{n}\text{P}_\text{R}\left(t\right)\space\text{d}t=\lim_{\text{n}\to\infty}\frac{1}{\text{n}}\cdot\int_0^\text{n}\text{I}_\text{R}^2\left(t\right)\cdot\text{R}\space\text{d}t=$$ $$\text{R}\cdot\left\{\lim_{\text{n}\to\infty}\frac{1}{\text{n}}\cdot\int_0^\text{n}\text{I}_\text{in}^2\left(t\right)\space\text{d}t\right\}\tag5$$

Now, the input current $$\\text{I}_\text{in}\left(t\right)\$$ that is stated in the integral at the end of equation $$\(5)\$$ can be found using the solution to the DE given in equation \$(4)$.

Question: How can I solve the integral given in equation $$\(5)\$$, that is the part where I do not get my head around?

EDIT:

I already found that, using the given values that:

$$\int_1^0\text{X}\left(t\right)\space\text{d}\tau=\frac{3}{8}\cdot\left(1-e^{800}\right)\tag6$$

And:

$$\frac{1}{\text{L}}\cdot\exp\left(-\frac{\text{R}}{\text{L}}\cdot t\right)=40e^{-800t}\tag7$$

And:

$$\int_1^t\text{X}\left(\tau\right)\space\text{d}\tau=\int_1^t\exp\left(\frac{\text{R}}{\text{L}}\cdot\tau\right)\cdot\left\{50+\frac{400}{\pi}\cdot\sum_{\text{n}=1}^\infty\frac{\sin\left(200\pi\cdot\text{n}\cdot\tau\right)}{\text{n}}\right\}\space\text{d}\tau=$$ $$\int_1^t\exp\left(\frac{\text{R}}{\text{L}}\cdot\tau\right)\cdot50\space\text{d}\tau+\int_1^t\exp\left(\frac{\text{R}}{\text{L}}\cdot\tau\right)\cdot\frac{400}{\pi}\cdot\sum_{\text{n}=1}^\infty\frac{\sin\left(200\pi\cdot\text{n}\cdot\tau\right)}{\text{n}}\space\text{d}\tau=$$ $$\frac{e^{800t}-e^{800}}{16}+\frac{400}{\pi}\cdot\sum_{\text{n}=1}^\infty\frac{1}{\text{n}}\cdot\left\{\int_1^t\exp\left(800\cdot\tau\right)\cdot\sin\left(200\pi\cdot\text{n}\cdot\tau\right)\space\text{d}\tau\right\}\tag8$$

Your approach is brave, and makes your life difficult. The first step to solving this problem is to realize that until the final calculation where you take the square of the current everything is linear.

Since that's true solve to find the contribution to I(t) from each term of the sum. (ignoring the DC component 50 for the moment)

$$Vin(t) = \frac{400}{\pi} \frac{\sin(200 n \pi t)}{n}$$ $$Vin(t) = R~I(t) + L \frac{dI(t)}{dt}$$

Solving that gives:

$$I(t) = 400\frac{200 L n \pi \cos(200 n \pi t) - R \sin(200 n \pi t)}{n \pi (40000~L^2 n^2 \pi^2 + R^2)}$$

Ignoring, WLOG, because we're after an average, any non periodic transient terms like $$e^{-\frac{R t}{L}}$$

You can now, using linearity, reassemble the sum

$$I(t) = 400\sum_{n=1}^\infty \frac{200 L n \pi \cos(200 n \pi t) - R \sin(200 n \pi t)}{n \pi (40000~ L^2 n^2 \pi^2 + R^2)}$$

One of the useful results you may, or may not, know from mathematics is that

$$\int_0^{2\pi} \cos(a t)\sin(b t)~dt =0, \forall a,b \in \mathbb{N}$$

and

$$\int_0^{2\pi} \sin(a t)\sin(b t)~dt =0, \forall a,b \in \mathbb{N} \text{ and } a\neq b$$

(similarly cos)

The integral you now need to do is

$$\int_0^1 400^2 \left( \sum_{n=1}^\infty \frac{200 L n \pi \cos(200 n \pi t) - R \sin(200 n \pi t)}{n \pi (40000~ L^2 n^2 \pi^2 + R^2)} \right) \left(\sum_{m=1}^\infty \frac{200 L m \pi \cos(200 m \pi t) - R \sin(200 m \pi t)}{m \pi (40000~ L^2 m^2 \pi^2 + R^2)} \right)dt$$

Using the relationships above you can quickly see that most of the terms in in the integral disappear, any sin multipled by a cos is zero, and any m not equal to n is zero. What you're left with is:

$$\int_0^1 400^2 \sum_{n=1}^\infty \frac{1}{(n \pi (40000~ L^2 n^2 \pi^2 + R^2))^2} (200 L n \pi \cos(200 n \pi t))^2 + (R \sin(200 n \pi t))^2 dt$$

pull the sum out through the integral and the remaining integrals are trivial.

$$\int_0^1 (cos(200 \pi n t))^2 dt = \int_0^1 (sin(200 \pi n t))^2 dt = \frac{1}{2}$$

I'll leave actually solving it, and thinking about what to do with DC component 50 for you to do, as it looks like a homework problem, and the preview is now too far away from the text editing window.

You might find the following standard result useful:

$$\sum_{n=1}^\infty\frac{1}{n(a^2+n^2)^2} = \frac{1}{4 a^4}\left(2\psi^{(0)}(1-ja)+2\psi^{(0)}(1+ja)+ja\psi^{(1)}(1-ja)-ja\psi^{(1)}(1+ja)+4\gamma \right)$$

Although that series converges exceedingly rapidly, so you may prefer a numerical solution.

Your approach and the suggest answer are awfully complex for a circuit with one singular load impedance $$\Z=R + jx\$$.

As this is homework, I can only guide you how I would approach this problem if I only had a pad of paper available (no Spice or Matlab).

1. Find the RMS value of each voltage source as a function of n.
2. Exploit super-position and write of the apparent power as a function of each voltage source (you will sum them together at the end).
3. The apparent power of the circuit is $$\V_{rms}^2/Z\$$.
4. Write out the load impedance as a function of n, $$\Z = R + j\omega L\$$, where $$\ \omega = 200 \pi n\$$.
5. Normalize the apparent power (divide top/bottom by complex conjugate of the denominator).
6. Power dissipated by the resistor is the real power. Leave the imaginary power out for recycling.
7. Complete the summation of the real power due to each voltage source.