Heating water

Since the resistor is in series with the battery, the equation

$\varepsilon = I (R+r)$

can be created.

Then, we can express the current $I$ as

$I = \frac {\varepsilon} {r+R}$ .

Then, since

$P = IV = I^{2}R$ ,

we can substitute $I$ to $\frac {\varepsilon} {r+R}$ by using the previous equation, to show $P$ , which shows the amount of power the resistor dissipated to the surroundings, can be represented as a function of $R$ since $\varepsilon$ and $r$ are constant.

Then,

$P = ( \frac {\varepsilon} {r+R} )^2 R = \frac {\varepsilon^2 R} {(r+R)^2} = \frac {144R} {(3+R)^2}$ .

After this point, if you do not know calculus, you can simply graph this equation into a calculator and find the maximum point. Then, the answer is $3.000 \Omega$ .

If you know calculus, you can differentiate this equation. To find the maximum power, we have to search the point where the value of $\frac {dP} {dR}$ must be zero or undefined. Then, by using the quotient rule,

$\frac {dP} {dR} = \frac {(3+R)^2 (144) - 144R(2(3+R))} {(3+R)^4} = \frac {-144 (R^2 - 9)} { (3+R)^4} = \frac {-144 (R-3)} {(3+R)^3}$ .

Then, the only points where $P$ is zero or undefined is when $R = 3$ , $R = -3$ , or $R= \infty$ However, since the resistance cannot be negative, the only possible critical point is when $R=3$ , $R = 0$ , and $R = \infty$ (0 can be a critical point since it is the endpoint of the interval of the resistance value, $[0 , \infty)$ .

However, the $R =0$ and $R = \infty$ are local minima, and the absolute maximum point is when $R=3$ .

Therefore, the resistance $R$ in Ohms that gives the maximum power dissipated in the resistor is when $R=3.000 \Omega$ .

The current through the load resistor is given by:

$i = \frac{E}{r+R}$

So, power dissipated in the load resistance is

$P = i^2 R = (\frac{E}{r+R})^2 R$

Now since for a particular source,the EMF $E$ & the internal resistance $r$ of the source are constant, $P$ depends on the value of R only,as asked in the problem, & $P$ is maximum when

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

or, $E^2 [\frac{(r+R)^2 - 2R(r+R)}{(r+R)^4}] = 0$

or, $(r+R)^2 - 2R(r+R) = 0$

or, $(r+R)(r-R) = 0.$

As $(r+R) \neq 0$ ,hence $r=R$ .

Back to our problem, $R = r = 3$ ohm.

Consider a battery, with EMF $\varepsilon$ and internal resistance $r$ . The battery is connected to a resistor. If the current drawn is $I$ , and the resistance of the connected resistor is $R$ , the voltage across the battery, and thus the resistor, $V$ , is given by the followng equation:

$V=\varepsilon-Ir$ -- (1)

Also, if the voltage across the resistor is $V$ , and the current drawn by the resistor is $I$ , the power drawn by the resistor, $P$ , is given by:

$P=IV$ -- (2)

By substituting (1) into (2):

$P=I(\varepsilon-Ir)$ -- (3)

By expanding (3), an equation relating the power dissipated by a resistor to the current flowing through that resistor can be found:

$P=\varepsilon I-I^2r$ -- (4)

To find what resistance gives the maximum power from a resistor, we need to find what volatge and current maximises power. By inspection, we can see that (4) is a negative quadratic of current. This means that the only place where $\frac {dP}{dI}$ is zero, is also the point at which power is at its greatest. Differentiating (4):

$\frac {dP}{dI}=\varepsilon-2Ir$ -- (5)

Substituting the values given in the question into (5), and setting $\frac {dP}{dI}$ to zero gives:

$0=12-(2 \times 3)I$

Thus giving a maximum current of $2A$ . By substituting this current, and the values given in the question into (1), the voltage that this current occurs at can be found:

$V=12-2 \times 3$

Thus giving a maximum voltage of $6V$ . By using the fact that $V=IR$ , $R$ can be found.

$R=\frac {V}{I}=\frac {6}{2}=3\Omega$

We know that V=IR. V=12 Volts and the equivalent resistance is R+3. P=\frac {I^2} {R}. I=\frac {12}{3+R}.So P=\frac {144R]{(R+3)^2}. So now we take the derivative of P with respect to R to get \frac {dP}{dR} = \frac {(R^2+6R+9)144-144R(2R+6)}{(R+3)^4}=\frac {-144(R^2-9)}{(R+3)^4}. Now we set that equal to 0 to get \frac {-144(R^2-9)}{(R+3)^4}=0 \rightarrow -144(R^2-9)=0 \rightarrow R^2-9=0 \rightarrow R^2=9 \rightarrow R=3.

We know that the internal resistance of r= 3 Ohms. Therefore,

$\frac{dP}{dR}$

= $\frac{dI^{2}R}{dR}$

= $\frac{d}{dR} \frac{V^{2}R}{(R + r)^{2}}$

= $V^{2} \frac{d}{dR} \frac{R}{(R + r)^{2}}$

= $V^{2} \frac{(R + r)^{2} - 2R( R + r )}{(R + r)^{4}} = 0$

Hence, from the numerator we can simplify the equation into

$R^{2} +2Rr + r^{2} - 2R^{2} - 2Rr = 0$

$r^{2}-R^{2} = 0$

$(r+R)(r-R) = 0$

Since resistance cannot have a negative value, $r-R=0$

$R= r = 3 Ohms$

This is a straightforward application of the maximum power theorem that states that maximum power transfer in a purely resistive circuit occurs when R = r. For the derivation see wikipedia , for example.

The power dissipated in the resistor $R$ (load) can be expressed as $P=I^{2}R$ where $I$ is current. This current can be found using Ohm's Law. It is simply the source voltage $\mathcal{E}$ over the total resistance $r+R$ . Thus the power-resistance dependence is given by the following function: $P(R)= {\mathcal{E}}^{2}\frac{R}{(r+R)^{2}}.$ The first derivative can be easily computed: $\frac{dP}{dR}={\mathcal{E}}^{2}\frac{r-R}{(r+R)^{3}.}$ From this expression we see that the function $P(R)$ has a critical point at $R=r=3~\Omega$ , so this point is either a maximum or minimum. Substituting back into the original expression for the power we see that it is indeed a maximum.

from Maximum Power Transfer Theorem in order to get the maximum power dissipated in the resistor

the resistor must equal the internal resistance

proof:

http://en.wikipedia.org/wiki/Maximum power transfer theorem#Calculus-based proof for purely resistive circuits

The Maximum Power Transfer Theorem basically states that to obtain the maximum power, a resistor must have the same resistance as the internal resistance of the circuit. As such, the answer is simply the given internal resistance $r = 3Ω$ . For the proof for this theorem, you can refer to this wikipedia article: http://en.wikipedia.org/wiki/Maximum power transfer theorem#Calculus-based proof for purely resistive circuits