# Maximum power transfer

## Homework Statement

Find L & C so that maximum power is delivered to load resistor.

___l 50 ohm l___ ( inductor L )____.____ ( inductor L )___
l . . . . . . . . . . . . .. . . . . . . . . . . l . . . . .. . . . . . . . . . l
10 V ac source . . . . . . . . ( Capacitor C) . . . . . . . l 200 ohm l
l______________________________l__________________l

Zi = Zl*

## The Attempt at a Solution

What bugs me is the fact that no source frequency is given.
Im guessing youd go about solving this by dividing the circuit into internal impedance and load impedance, lets pick:

Zi = 50 + jwL
So that the load circuit consists of the cap, other inductor and 200ohm resistor.
The cap is connected in parallel with the inductor and resistor so
1/(jwC) ll (200 + jwL)
((200 + jwl)/(jwC))/(jwC + 200 + jwL)
equating that to Zi*
((200 + jwl)/(jwC))/(jwC + 200 + jwL) = 50 -jwL

Is there any sense to this?

Ps: sorry for the dingy drawing.

berkeman
Mentor

## Homework Statement

Find L & C so that maximum power is delivered to load resistor.

___l 50 ohm l___ ( inductor L )____.____ ( inductor L )___
l . . . . . . . . . . . . .. . . . . . . . . . . l . . . . .. . . . . . . . . . l
10 V ac source . . . . . . . . ( Capacitor C) . . . . . . . l 200 ohm l
l______________________________l__________________l

Zi = Zl*

## The Attempt at a Solution

What bugs me is the fact that no source frequency is given.
Im guessing youd go about solving this by dividing the circuit into internal impedance and load impedance, lets pick:

Zi = 50 + jwL
So that the load circuit consists of the cap, other inductor and 200ohm resistor.
The cap is connected in parallel with the inductor and resistor so
1/(jwC) ll (200 + jwL)
((200 + jwl)/(jwC))/(jwC + 200 + jwL)
equating that to Zi*
((200 + jwl)/(jwC))/(jwC + 200 + jwL) = 50 -jwL

Is there any sense to this?

Ps: sorry for the dingy drawing.

I would interpret the source impedance as just the 50 Ohms. The rest of the elements form the load. You need to pick the L and C values to match the real 50 Ohms, I would think.

You might then see if any changes to the LC values could deliver more power to the actual load resistor... I would think not, but you should definitely check. If the load circuit it matched to the source 50 Ohms, then all power should go forward, and the only place for it to be dissipated is in the load resistor. (EDIT -- well half of the source's power is dissipated in the source resistor, but the rest should get dissipated in the load resistor as long as the network is matched)

This may also look a lot like using a T network to match a different source and load resistance. You could use the math of matching networks to double-check your answer...

I would interpret the source impedance as just the 50 Ohms. The rest of the elements form the load. You need to pick the L and C values to match the real 50 Ohms, I would think.

You might then see if any changes to the LC values could deliver more power to the actual load resistor... I would think not, but you should definitely check. If the load circuit it matched to the source 50 Ohms, then all power should go forward, and the only place for it to be dissipated is in the load resistor. (EDIT -- well half of the source's power is dissipated in the source resistor, but the rest should get dissipated in the load resistor as long as the network is matched)

This may also look a lot like using a T network to match a different source and load resistance. You could use the math of matching networks to double-check your answer...

Since it says 10V ac, you may assume that the source frequency is 60 Hz.
I agree with Berkeman. You should calculate what values of L and C present to the primary an impedance of 50 ohm and to the secondary an impedance of 200 ohm.