Electronic – How to calculate the phase difference between current and voltage

circuit analysisfourierpower electronics

Using the next circuit and the components of

schematic

simulate this circuit – Schematic created using CircuitLab

  • \$R=20\Omega\$

  • \$L=25mH\$

  • \$V_{DC}=36\$

  • \$v_{s}(t)=50+\underset{n=1}{\overset{\infty}{\Sigma}}(\frac{400}{n\pi})sin(200n\pi t)\$

The idea is determinate the average power absorbed by R.

So using the Fourier series based on the generalized Parseval identity
\$\frac{1}{L}\int_{0}^{c+2L}f(x)g(x)dx=\frac{a_{0}c_{0}}{2}+\overset{\infty}{\underset{n=1}{\Sigma}}(a_{n}c_{n}+b_{n}d_{n})\$

\$P=V_{o}I_{0}+\underset{n=1}{\overset{\infty}{\Sigma}}\frac{V_{m}I_{m}}{2}cos(\theta_{n}-\phi_{n})\$

\$I_{0}=\frac{V_{0}-V_{DC}}{R}=\frac{50-36}{30}=0.7\,A\$

\$P_{0R}=(0.7^{2})(20)=9.8\$
\begin{array}{|c|c|c|c|c|c|}
\hline
n & V & I & Z & Angle & P\\ \hline
\hline
\hline
0 & 50 & 0.7 & 20 & 0 &9.8\\ \hline
\hline
1 &\frac{400}{\pi}=127.32 & \frac{V}{Z}=5.00 & \sqrt{20^{2}+(200\pi25mH)^{2}}=25.43 & & \\ \hline
\hline
2 &\frac{400}{2\pi}=63.66 & 1.7& 37.24 & & \\ \hline
\hline
3 & \frac{400}{3\pi}=42.44 & 0.829 & 51.19 & & \\ \hline
\hline
4 & \frac{400}{4\pi}=31.83 & 0.4827 & 65.93& & \\ \hline
\hline
5 & \frac{400}{5\pi}=25.46 & 0.3141& 81.04 & & \\ \hline
\hline
\end{array}

But where Im stuck is how can be calculated the phase difference (the angle column) between the current and the voltage, since I think the second angle of \$cos(\theta_{n}-\phi_{n})\$ would be zero for every term of the series.

UPDATE
Whith the idea of The Photon that the inductor current indeed is out of phase but this can't cause the resistor voltage and current to be out of phasem and using the RL impedance

\begin{array}{|c|c|c|c|c|c|}
\hline
n & V & I & Z & Angle & P\\ \hline
\hline
\hline
0 & 50 & 0.7 & 20 & 0 &9.8\\ \hline
\hline
1 &\frac{400}{\pi}=127.32 & \frac{V}{Z}=5.00 & \sqrt{20^{2}+(200\pi25mH)^{2}}=25.43 & 0.665 & 250.47 \\ \hline
\hline
2 &\frac{400}{2\pi}=63.66 & 1.7& 37.24 & 1.003 & 29.099\\ \hline
\hline
3 & \frac{400}{3\pi}=42.44 & 0.829 & 51.19 & 1.169& 6.879\\ \hline
\hline
4 & \frac{400}{4\pi}=31.83 & 0.4827 & 65.93& 1.262 &2.334 \\ \hline
\hline
5 & \frac{400}{5\pi}=25.46 & 0.3141& 81.04 & 1.321 &0.988 \\ \hline
\hline
\end{array}

So the average power absorbed by R is \$299.57 W\approx300 W\$

Best Answer

Hints:

  • You can use the impedance of the R-L combination to find the current through the circuit due to each Fourier component.

  • You can then use Ohm's law to get the voltage across the resistor due to each Fourier component.

  • Then you can use the formula for power (derived from Parseval's theorem) as stated in your post to get the power.

One more hint:

  • Since the inductor, on average, consumes no power, the average power dissipated by the resistor is equal to the average power delivered by the sources in this circuit. So after you've found the circuit current, you might rather just calculate the source power than the resistor power.