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A 12V battery has an internal resistance of $2.0\Omega$. A load of variable resistance is connected across the battery and adjusted to have resistance equal to that of the internal resistance of the battery. Find the power dissipated.

The equation for internal resistance is $$\mathcal{E}=I(R+r)$$ where $R$ is the resistance in the circuit, $r$ is the internal resistance of the battery and $I$ is the current. We know that $\mathcal{E}=12V$. The sum of internal and external resistance would be $4 \Omega$, seeing as the variable resistor is set to $2 \Omega$, which adds to the already exisiting $2 \Omega$ from the battery.

The equation for power dissipated is

$$P=I^2R=\frac{V^2}{R}$$

So I substitued the value for $V$ and the value for $R$ to get

$$P=\frac{12^2}{4}=36$$

This is, however, wrong. Any ideas as to why?

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  • $\begingroup$ If $36\,\rm W$ is wrong then the setter of the questioned possibly wanted the power dissipated just in the external resistor. $\endgroup$ – Farcher Dec 8 '18 at 15:52
  • $\begingroup$ @Farcher That would make sense, but then wouldn't the power be $\frac{12^2}{2}$? The OP says the answer is $18\ \rm W$ in another comment $\endgroup$ – Aaron Stevens Dec 8 '18 at 16:46
  • $\begingroup$ @AaronStevens No, it would be a half of the total power dissipated ie the total power is shared between the internal resistor and the external resistor. $\endgroup$ – Farcher Dec 8 '18 at 16:49
  • $\begingroup$ @Farcher But if you have a resistor $R$ with a potential drop of $V$ across it, then isn't the power dissipated in the resistor $\frac{V^2}{R}$, regardless of what else is going on in the circuit? Is the potential drop across the external resistor not $12\ \rm V$? $\endgroup$ – Aaron Stevens Dec 8 '18 at 16:56
  • $\begingroup$ @AaronStevens There is 6V across each of the resistors to make the 12 V generated by the battery. $\endgroup$ – Farcher Dec 8 '18 at 17:00
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You aren't applying your power equation correctly. You can't just plug in any voltage, current, or resistance.

The equation $P=IV$ gives you the power delivered or dissipated by a single part of the circuit with a potential drop $V$ across it and current $I$ flowing through it.

If you want the total power dissipated in the circuit, you must apply this to each element separately. If just the outer resistance, then only consider this resistor. I will leave this to you.

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  • $\begingroup$ So would I need to add the power dissipated in the battery to the power disspiated in the circuit? If so, what's the current across the circuit? $\endgroup$ – Pablo Dec 8 '18 at 15:08
  • $\begingroup$ @Pablo Use ohms law on the variable resistor $\endgroup$ – Aaron Stevens Dec 8 '18 at 15:27
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Why do you substitute $R$ with 4? $R$ "is the resistance in the circuit", so it equals $2\Omega$, not $4\Omega$. There is another question though: do they need total dissipation or just dissipation in $R$?

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Your answer looks right to me as long as you want the total power dissipated (including in the battery).

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  • $\begingroup$ What is the given answer? 18 W? $\endgroup$ – Ben51 Dec 8 '18 at 14:56
  • $\begingroup$ Yes, but $\frac{12^2}{4}=36$. How did you get to $18$? $\endgroup$ – Pablo Dec 8 '18 at 15:09
  • $\begingroup$ 18 W dissipated in the resistor, 18 in the battery $\endgroup$ – Ben51 Dec 8 '18 at 17:01

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