power supplied to a resistive load cannot exeed (V\2)^2/R^2

• Oct 26th 2010, 12:34 PM
bigwave
power supplied to a resistive load cannot exeed (V\2)^2/R^2
A battery has internal resistance $\displaystyle R_i$

and open circuit terminal voltage $\displaystyle V_b$

Show that the power supplied to a resistive load cannot exceed $\displaystyle \frac{({\frac{V_b}{2})^2}}{R_i}$

well I only changed this to

$\displaystyle \frac{{\frac{V_b}{2}^2}}{R_i} \Rightarrow ({\frac{V_b}{2})^2}\times\frac{1}{R_i} \Rightarrow \frac{V_b^2}{4R_i}$

but not sure how you determine what cannot be exceeded

any suggest...(Cool)

mod.... this should of been put in "other topics" its not an advanced topic.
• Oct 26th 2010, 01:00 PM
zzzoak
Internal resistance r.
External R.

$\displaystyle \displaystyle { I=\frac{V}{r+R} }$

$\displaystyle \displaystyle { P=I^2 \; R \; = \; \frac{V^2}{(r+R)^2} \; R }$

Max power on R is

$\displaystyle \displaystyle { \frac{dP}{dR} \; = \; 0 }$

we get

$\displaystyle R=r$

and

$\displaystyle \displaystyle { P(max)=\frac{V^2}{4r} }$
• Oct 26th 2010, 01:10 PM
Ackbeet
I've attached a circuit model of the situation. Would you agree with this model?

Attachment 19481

Let's say this model is accurate. Now what? Well, the one constant thing is the voltage of the battery. That doesn't change (at least, not theoretically). But as you change the load resistance, the relative voltage drops across each resistor changes. You want to maximize the power in the load. Now, Ohm's Law yields

$\displaystyle V_{\text{b}}=I(R_{\text{i}}+R_{\text{load}}).$

The power dissipated in the load resistor is $\displaystyle P_{\text{load}}=I^{2}R_{\text{load}}=\left(\dfrac{ V_{\text{b}}}{R_{\text{i}}+R_{\text{load}}}\right) ^{2}R_{\text{load}}.$

Now use the usual Calc I procedure of setting

$\displaystyle \dfrac{dP_{\text{load}}}{dR_{\text{load}}}=0,$ and solve for $\displaystyle R_{\text{load}}.$

What do you get?
• Oct 26th 2010, 01:29 PM
bigwave
verify
sure appreciate the help... now I see how this works...thanks...

there was a follow up question on this.... not sure if input it right

thus if $\displaystyle V_b=96V$ and $\displaystyle R_i = 50\Omega$.

Discrete loads of $\displaystyle 150, 100, 50, 30,$ and $\displaystyle 20m\Omega$ are connected, one at a time, across the battery. Plot the curve of power supplied versus the ohmic value of the load. I tried to use the Wolfram|Alpha to graph this but didn't get the expression right http://www.wolframalpha.com/input/?i=plot[P%3D%28%2896%29^2%29%2F%284\Omega%29%3B%28P%2C-50%2C200%29%2C%28\Omega%2C-50%2C200%29]

Hence, verify that the maximum power transfer occurs when $\displaystyle R_i = R_{load} = 50\Omega$
• Oct 26th 2010, 05:13 PM
Ackbeet
Well, here's a continuous plot. If you wanted to do a discrete plot, I'd use Excel.
• Oct 26th 2010, 07:59 PM
bigwave
appreciate the plot

well. I need a lot of help with this EE subject.

more ?? to post on the way
• Oct 27th 2010, 02:02 AM
Ackbeet
Did you mean you need more help with this problem, or just more help with problems in the same subject?
• Oct 27th 2010, 02:23 PM
bigwave
Im ok with this one

but I will be posting more new problems

its a new subject to me.
• Oct 27th 2010, 04:16 PM
Ackbeet
Ok. I'll be looking for more threads. Have a good one!