I was given a problem, the conditions are: The voltage and current for a device (using the passive sign convention) are periodic functions with \$T=100 ms\$ described by

\$v(t)=\begin{cases} 10\,V & 0\,ms<t<70\,ms\\ 0 & 70\,ms<t<100\,ms \end{cases}\$

\$i(t)=\begin{cases} 0\,A & 0\,ms<t<50\,ms\\ 4\,A & 50\,ms<t<100\,ms \end{cases}\$

So the questions from the book are instantaneous power, the average power, and the energy per period, but I have no interest in these, What I really want to know is if a DC source is connected to the circuit , how can be calculated the average power? a.k.a. What would be the average current (\$I_{avg}\$)?

I understand that the \$I_{avg}\$ would be the given current over the period

\$I_{avg}=\frac{4 A}{100 ms}=0.00004 A\$

  • 4
    \$\begingroup\$ Average would be the integral over time period divided by that time period. \$\endgroup\$ – Eugene Sh. Dec 17 '18 at 17:05
  • 2
    \$\begingroup\$ This same question was asked previously in the past 48 hours. What did you learn from the comments you got when you asked before? \$\endgroup\$ – The Photon Dec 17 '18 at 17:07
  • 1
    \$\begingroup\$ I'm not sure how far off you are, but I will give you a hint, the answer starts with the number 2. \$\endgroup\$ – Harry Svensson Dec 17 '18 at 17:16

Thanks to Eugene Sh

$$i(t)=\frac{1}{T}\int i(t)dt=\frac{1}{100\times10^{-3}}(\int_{0}^{50\times-3}0+\int_{50\times-3}^{100\times-3}4)dt=2\,A$$


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.