# Why is the R resistance equal to the internal resistance

The answer to the attachment is if the R resistance was equal to the internal resistance. How was this determined? Please explain this to me because i have no idea how they can solve this. Thanks

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anyone have any idea as to how the answer was reached?

I haven't taken any courses on circuits yet and only vaguely remember what I covered in intro. But here's what I think.

The potential of the battery and the current produced are both constant.

Since the only components are the battery and the resistor, the potential drop across the resistor must equal the potential provided by the battery. So the maximum value of R for which the circuit would "function" is when it is equal to the internal resistance.

It = IR where t = internal resistance.

GCT
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latex

From what I recall recently

$$V_{battery} = emf-IR_{internal}$$

assuming that the "load" resistor is $$R_{eq}$$, (I forgot some of the terminologies), than

$$P=I^{2}R$$, where the $I=EMF/(R+R_{internal})$

thus

$$P= \frac {(EMF^{2}(R))}{(R+R_{internal})^{2}}$$
Try observing the trend for an infinitely large value of R and infinitely small value.

I'm pretty new to this also, so I'm not absolutely certain

I once derived this as an out of class exercise:

$$P = \frac{E^2R}{(R+r)^2}$$ as shown above.

where P is the power, R is the load, r is the internal resistance and E is the battery EMF.

We are trying to find the point at which P is a maximum by varying the load R, therefore we involve a bit of calculus with respect to R.

We need the quotient rule.

$$\frac{d}{dx}(\frac{u}{v}) = \frac{V\frac{du}{dx}-U\frac{dv}{dx}}{V^2}$$

letting $$u = E^2R$$ and $$v = (R+r)^2$$

So ....

$$\frac{dP}{dR}=\frac{E^2(R+r)^2 - 2(R+r)(E^2R)}{(R+r)^{4}}$$

at maximum power $$\frac{dP}{dR}=0$$
$$(R+r)^{4}$$ can not equal zero, therefore $$E^2(R+r)^2 - 2(R+r)(E^2R)$$ must equal zero.

Rearranging this eventually gets to r = R, thus showing at maximum power the load resistance equals the internal resistance