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I have a question that asks me to calculate the maximum power output from a power supply that can be delivered to a load.

I know what the answer is (0.5W) but I am struggling to get too it. I tried converting the Voltage source \$V_{RMS}\$ to \$V_{Max}\$ by multiplying by \$\sqrt{2}\$ and then using this value to calculate \$I_{Max}\$ then by applying \$P_{Max}=I_{max}V_{Max}\$ I get a value of 4W which is wrong. I think I need to factor in the frequency \$F\$of the voltage source to calculate the power however I am unsure of the equation I need, without knowing time \$t\$. Is there a specific method or equation I should know about?

The load in the circuit diagram is used in the second part of the question, and I don't think it is needed in this first part.

Question Circuit Diagram

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  • \$\begingroup\$ To be sure, I'm certain the question is asking for the maximum average power \$P\$, not the maximum instantaneous power \$p(t)\$. This an AC (phasor) problem, not a time domain problem so there is no \$t\$ to be concerned with. \$\endgroup\$ – Alfred Centauri Jun 7 '14 at 20:28
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If you know that maximum power transfer happens when the impedances are matched (complex conjugate if they're not real), then maximum power is when the load is 50 ohms (real) you can simply calculate P = \$ \dfrac { V_{RMS}^2}{R}\$ = 1W total, but only half of that is in the load, so 0.5W.

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According to Wikipedia - Maximum power transfer theorem, max power delivered to the load will be when load resistance becomes equal to the source resistance, so for load = \$50 \Omega\$, current = \${10 \over (50+50)}=0.1A\$ and power delivered to load = \$0.1^2 \cdot 50 = 0.5W\$

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