Engineering Calculate the maximum power of the resistor in a circuit

[diredragon]

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?

 

[scottdave]

The current is the same theough all elements, since they are all in series.

 

[NascentOxygen]

My question is since we are looking for Pp=RpI2Pp=RpI2P_p = R_pI^2, the active power, shouldn't RpRpR_p equal the real part of Z and not it's effective value?

Try both loads and determine which has the higher load power. Or, you could set yourself a numerical exercise to test this. Picture a source with impedance Z=1+j1. Is there more power in the load if resistance of load is 1Ω or √2Ω? Calculate for both cases.

 

[scottdave]

You need to consider the entire Z, to know how much current is passing through the resisitor. As @diredragon said, try different values. If you have a way to plot this, then plot over a range of values.

 

[scottdave]

Also, the resistor doesn't "care" what the phase of the current running through it is. The voltage across a resistor will be in phase with the current passing through it, so you can just multiply the (magnitude of the current)² by the resistance, to get the power for that resistor.

As for Average Power, think of it as the power generated by the portion of current which is in-phase with the voltage. So if you look at the entire circuit, then just look at the real portion of the current delivered, that will be the amount which is in phase with the source voltage.

 

[scottdave]

For these types of problems, a calculator, which handles complex numbers, really saves the day. If you don't have that, then perhaps you have access to MATLAB or Maple. I suppose Wolframalpha could do the work, too. Another thing to look at: Python (a programming language) can handle complex numbers. It might be worth a look, since it is free.

 

[Babadag]

I agree with scottdave it would be better if you could use a program handling complex numbers-in my opinion even Microsoft Excel (from 2010 edition) would do the job. However it is still not so complicate if you will insert all the resistance in the circuit in order to calculate the current. And then the power dissipated in circuit will be P=(Rg+Rp)*I^2. Now you have to put d(P)/d(Rp)=0. The result will be Rp=2*w*Lg-Rg I think for the maximum power.

 

[Babadag]

If R in b) it is Rp then P(Rp)=Rp*E^2/((Rg+Rp)^2+Xg^2) and from dP/dRp=0 (for maximum) you'll get Rp=sqrt(Rg^2+Xg^2) =100 ohm indeed.
Now "the average power" in this circuit has to be total losses in Rp and Rg and then the result is 0.18 W indeed.
The complication is not in the calculation just in the logic.

 

[diredragon]

Thanks for the replies. I do get it now :)