- #1
diredragon
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Homework Statement
In an AC circuit of periodic current we know: ##E=6[V]##, ##R_g=50[Ω]##, ##L_g=\sqrt3 [nH]## and ##ω=50*10^9[s^{-1}]##. Calculate:
a) All available power of the ##E##
b) The resistance of ##R## so that the average power is max
c) That average power from part b)
Homework Equations
3. The Attempt at a Solution [/B]
First the equivalent resistance of the left side is
##Z = R_g + jωL_g = 50(1+j\sqrt3)##.
For the part b)
My question is since we are looking for ##P_p = R_pI^2##, the active power, shouldn't ##R_p## equal the real part of Z and not it's effective value? If not, in which case is that the solution since I've seen a couple of problems having that solution. I am not sure if they had ##R_p## or some ##Z_p## on the right.
The book gives the answer ##R_p = \left | Z \right |## = 100.
For the part c)
In this case the average power ##P_p=R_p* \frac{E^2}{(|Z + R_p|)^2} = 100*\frac{36}{(150+j50\sqrt3)^2}=120[mW]##
For the part a)
I don't know what all available means but i supposed it's the power of the generator as it is in this resistance but it's not.
The solution for c) is ##\frac{E^2}{4R_g} = 180 [mW]##. How to get this?