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I am badly stuck in finding value for R in this circuit for maximum power transfer in the network, enter image description here

It is my assignment question and I am badly confused.

Please help , Thanks.

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    \$\begingroup\$ You need to show your effort at solving this problem/homework. Read the FAQ. \$\endgroup\$ – winny Oct 16 '16 at 13:09
  • \$\begingroup\$ Reduce to a thevanin equivalant circuit \$\endgroup\$ – JIm Dearden Oct 16 '16 at 13:15
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For maximum power transfer you are expected to not only know how to set up equivalent circuits, but also how to find maximums and minimums by setting a derivative to zero.

The first step, of course, is to develop the equivalent minimum circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

You should have been able to follow the above steps. If not, then you have some serious study to do and you won't yet be able to solve these problems until you can perform the above steps on your own.

Now that you have the final circuit, you can work out the following equation:

$$\begin{align*} I &= \frac{V}{R+R_3} \\ \\ P_3 &= I^2 R_3 = \frac{V^2 R_3}{\left(R+R_3\right)^2} \end{align*}$$

Now, in the above equation for \$P_3\$, \$V\$ and \$R\$ are known constants. \$R_3\$ is the variable whose value you want to find in order to maximize \$P_3\$. This is the point where you are supposed to know from calculus courses (or reading) that you take the derivative, set that new expression to zero as an equation (to find where the slope is zero), and then solve. This will give you either a minimum or a maximum.

$$\begin{align*} \frac{\textrm{d}P_3}{\textrm{d} R_3}&=V^2\frac{R-R_3}{\left(R+R_3\right)^3} \\ \\ \textrm{so,} \\ \\ V^2\frac{R-R_3}{\left(R+R_3\right)^3}&= 0 \end{align*}$$

I'll leave the last step of solving that answer to you. But that's all that is left.

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when you're RL = Rth you get maximum power in RL first calculate RL :

Rth=(50||50) + (40||60) then Rth = 25 + 24 = 49 ohm

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    \$\begingroup\$ You shouldn't simply give answers to homework suggestions. I recommend deleting this answer, or at least editing it, so that the OP has to do at least some of the work himself. \$\endgroup\$ – DerStrom8 Oct 16 '16 at 14:36
  • \$\begingroup\$ @arashzgh Does OP know why there is maximum power when Rl = Rth. DerStrom8 is right. Help OP to understand. This way he is only learning to trick himself and others \$\endgroup\$ – Decapod Oct 16 '16 at 16:00

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