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

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

A battery has internal resistance $R_i$

and open circuit terminal voltage $V_b$

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

well I only changed this to

$\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...

mod.... this should of been put in "other topics" its not an advanced topic.

2. Internal resistance r.
External R.

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

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

Max power on R is

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

we get

$
R=r
$

and

$
\displaystyle { P(max)=\frac{V^2}{4r}
}
$

3. I've attached a circuit model of the situation. Would you agree with this model?

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

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

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

Now use the usual Calc I procedure of setting

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

What do you get?

4. ## 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 $V_b=96V$ and $R_i = 50\Omega$.

Discrete loads of $150, 100, 50, 30,$ and $20m\Omega$ 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 $R_i = R_{load} = 50\Omega$

5. Well, here's a continuous plot. If you wanted to do a discrete plot, I'd use Excel.

6. appreciate the plot

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

more ?? to post on the way

7. Did you mean you need more help with this problem, or just more help with problems in the same subject?

8. Im ok with this one

but I will be posting more new problems

its a new subject to me.

9. Ok. I'll be looking for more threads. Have a good one!