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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\$

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    \$\begingroup\$ Average would be the integral over time period divided by that time period. \$\endgroup\$ – Eugene Sh. Dec 17 '18 at 17:05
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    \$\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
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    \$\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
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UPDATE
Thanks to Eugene Sh

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

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