# Max power

Discussion in 'Homework Help' started by boks, Nov 2, 2008.

1. ### boks Thread Starter Active Member

Oct 10, 2008
218
0

L last = 1.0 H
R last = 100 Ohm
R = 10 Ohm
V rms, s = 100/sqrt(2) V
f = 400 Hz

What component should be placed between A and B to maximize the effect in R last and L last (last = load)? What's the maximum effect in R last?

We have maximum power when Z circuit = Z load. R load is the real part of Z load. I can't see how to make Z circuit = Z load when R load is different from R...

2. ### mik3 Senior Member

Feb 4, 2008
4,846
63
Z=a+jb

where

a=resistance
b=reactance, in this case is inductive reactance

Zcircuit=(R+a)+jb until now
and

for these two impedances to be equal

(R+a)=100 and b=XL

Thus you have to place a 90 ohm resistor in series with a 1H inductor to have maximum power on the load.

3. ### vvkannan Active Member

Aug 9, 2008
138
11
But should not the source and load impedences be complex conjugates for the power to be maximum?

4. ### mik3 Senior Member

Feb 4, 2008
4,846
63
If the inductor was not there, then to have maximum power on Rload, you should place a short circuit across A and B. However, you have an inductor and you want to eliminate the voltage the voltage drop across it to have maximum power transfer on Rload. To do it you need to connect a capacitor across A and B with the same reactance as the inductor at 400 Hz. Thus, you are partially right Vvkannan and i apologise to Boks for this mistake, sometimes you cant see the right solution from the first time!!