# Max power transfer

Discussion in 'Homework Help' started by minduka, Mar 8, 2009.

1. ### minduka Thread Starter New Member

Mar 5, 2009
2
0
Hi

I need some help on two max power transfer questions.

This is my approach to the first one.

$R_{L}= 8 \Omega$ - because max power occurs when $R_{TH}$ = $R_{L}$

Then I found the max power transfered in (b) by using

$P_{L,Max} = \frac{v_{L}^{2}}{4R_{TH}}$

$V_{L} = IR = 2A*8\Omega = 16v$

$R_{TH} = 16//8//16 = 4\Omega$

$P_{L,Max} = \frac{16^{2}}{16} = 16$

I'm confused when they give you $R_{L}$ and need to find the max power. I'm assuming you have to use $R_{L}$ to find the max power.

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For the second problem:

$R_{TH} = 10//10+10 = 15\Omega$

$V = 20v$

$P = \frac{20^{2}}{4*15} = 6.67 W$

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Hello minduka,

In the first question you recognized that RL = 8Ω for the left hand circuit.

The questioner is not asking for the maximum power transfer case for the right hand circuit - they are asking you to place the load RL = 8Ω into the right hand circuit and then solve for the power dissipated in the load, which with RL = 8Ω would give 32W - as indicated by the correct answer.

In the second question, you are nearly there, but just need to realize that the voltage you should use is not 20V. Suppose you reduce the circuit to its Thevenin equivalent. It would be a 10V source in series with a 15Ω resistance. So the maximum power case is ((10/2)λ2)/15 = 1.67W.

Hope this helps.

3. ### minduka Thread Starter New Member

Mar 5, 2009
2
0

I think the first problem is a little decieving

4. ### nirosha New Member

Nov 6, 2008
1
0
In the first problem,by using voltage division rule we get voltage across RL.use this voltage and given 2watts to find RL then place this RL value in the next ckt and solve the rest.

5. ### Ratch New Member

Mar 20, 2007
1,068
4
minduka,

Not necessily so. If the source impedance and voltage cannot be changed, then yes, max power occurs when the load impedance equals the source impedance. But if you can make the source impedance smaller while keeping the source voltage the same, then you will transfer more power with the smaller source impedance.

Ratch