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For all problems, assume that the battery voltage and lamp resistances are constant, no matter what current is present.

A 75-W lamp is connected to $125 V$.

  • What is the current through the lamp?

$$I=\frac{75}{125}=\frac{3}{5}=0.6A$$

A resistor is added to the lamp in the previous problem to reduce the current to half of its original value.

  • What is the potential difference across the lamp?

I think that the potential difference across the lamp is just 125V. But in the solutions manual the answer is $6.3×10^1V$ $$V=I \cdot R=0.3 \cdot 2.1 \times 10^2=6.3 \times10^1 V$$.

I want to know why they use the resistance of the lamp before the change $(2.1×10^2)$ instead of the resistance after the change $(4.2×10^2)$?

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closed as off-topic by John Rennie, BMS, Brandon Enright, ACuriousMind, Kyle Oman Nov 6 '14 at 6:30

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It depends on what is meant by "added to the lamp". If the resistor is in series with the lamp, then there will be a voltage drop across that resistor (since a current is flowing) and less of the potential is available for the lamp. But if the lamp is modified so it has higher resistance (a perfectly acceptable interpretation of the wording) then the lamp is still the only thing in the circuit and it has the same potential across it.

Note that the resistance of a lamp is a strong function of temperature, and when you drop the current in half with a series resistor, you probably changed the temperature of the filament and thus the resistance. This means that the "standard answer" is wrong. But it continues to be taught...

In pictures:

enter image description here

If $R \ne R'$ then the resistor that is added will be $2R-R'\gt R$ in order to cut the current in half - and the voltage across the lamp (now with resistance $R'$) will be less than half the potential $V$, namely $V_{lamp}=\frac{i}{2}R' < \frac{i}{2} R = V_{original}/2$.

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  • $\begingroup$ Sorry, I forgot to put the note, Could you take a look at the question again, Floris? $\endgroup$ – Wael Alsafi Nov 5 '14 at 19:16
  • $\begingroup$ OK I see the update. Now $R' = R$, but otherwise the diagram is the same. The only thing you need to ask is "is the series resistor considered part of the lamp, or in series with the lamp". Because if the resistance of the lamp is still $R$ (rather than $2R$), then voltage across the lamp is half. But if the additional resistance is part of the lamp then you still have the full voltage across the "lamp". Does that make sense? $\endgroup$ – Floris Nov 5 '14 at 19:55
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If you add a resistors in series to another resistor (lamp), the voltage across the lamp will decrease. The new voltage will be:

$V_{LAMP}=R_{lamp}V_{total}/(R+R_{lamp})$

The resistance in the lamp is a constant property. Regardless of the change the resistance is constant, it is the voltage across it that's different.

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