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I just learnt how potentiometer circuits work, and I was taught that the resistance in the secondary circuit does not change the balance length as no current flows through the secondary circuit and thus the only potential drop is through the EMF of the unknown battery in the secondary circuit. However my proffessor mentioned that if the resistance of bulb x was decreased in this particular arrangement, the balance length would shorter!

The reason he gave was "This is because with a lower resistance bulb used for X, the current flowing in the lower circuit increases, the voltage drop across the internal resistor increases, and hence the terminal potential difference across the cell in the lower circuit decreases. " However I am struggling to understand why the concept of no current flowing through the circuit does not apply here, I still feel that decreasing the resistance of bulb x would not affect the balance length for the reason stated above. Thank you for clarifying this conceptual error of mine in potentiometer circuits! here is the circuit

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However I am struggling to understand why the concept of no current flowing through the circuit does not apply here

The condition required for there to be zero current through the ammeter connected to $C$ is this: all of the current through the bulb $Y$ is through the battery labeled with voltage $E$ (note that the battery is modeled as an ideal battery in series with an internal resistance).

If you'll think about this a little bit, you'll realize that if there's zero current through the ammeter, the ammeter can be removed with changing the value of any voltage across or current through in the circuit (if that isn't clear, stop here and think about it some more).

Now, since the (ideal) ammeter has zero volts across (regardless of the current through), it follows that with the ammeter in circuit, the voltage AC equals the voltage across the cell.

So, stipulate that the potentiometer is adjusted such that the ammeter reads zero and then the ammeter is removed. Given the above, it must be that the voltage AC is still equal to the voltage across the cell.

That is, the condition that there is zero current through the ammeter is the condition that the voltage AC equals the voltage across the cell when the ammeter is not connected.

Carefully note that this does not imply that there is no current through the bulbs!

if these bulbs are in parallel with the battery, musn't their potential drops match that of the battery in the secondary circuit?

Correct (if you mean the combination of the bulbs and not the individual bulbs), but that potential drop (voltage across) isn't constant, it is a function of the resistance of the bulbs, the internal resistance of the cell, and the voltage $E$. Assuming the ammeter has been removed, the voltage across the cell is given by:

$$V_{cell} = E\left(1 - \frac{r}{r + R_Y + R_X || R_Z}\right)$$

where

$$R_X || R_Z = \frac{1}{\frac{1}{R_X} + \frac{1}{R_Z}}$$

Only in the case that $r=0$ is this a constant.

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  • $\begingroup$ Thank you @AlfredCentauri, you have clarified much about the "zero current" concept, if I'm understanding you correctly, there is still current flowing through the bulbs (provided by the battery). However regarding your last equation, I recognise it as the expression of the terminal voltage of a cell. Would I be right to say that the terminal voltage across the cell in the secondary circuit is equal to the voltage drop across the balance length? If not, I am still unsure about the concept of how to find the new balance length if I were to add additional resistors in parallel to the 2nd cell. $\endgroup$ – Lucas Tan May 11 '20 at 22:54
  • $\begingroup$ As I thought about the concept of how zero current through the ammeter simply implies that if I were to remove the ammeter, nothing is affected, I have a query about it. It is this: does current flow through the wire connecting the main circuit? If so (or if not), what is the source of current of the bulbs? At first I thought it would be from the driver cell, however on second thoughts I realised the current source could be from the secondary cell, or both. So where would the source of current be for the bulbs to light up, thank you! $\endgroup$ – Lucas Tan May 11 '20 at 23:16
  • $\begingroup$ @LucasTan. When the ammeter is removed, there is a current circulating in the 'upper' circuit (6V battery, resistor and pot) as well as the 'lower' circuit (E volt battery and bulbs). Adjusting the pot will not change the current in the upper circuit (the wiper terminal is disconnected) but it will change the voltage across AC. Changing the bulb resistance will change the voltage across the cell in the lower circuit. If the voltage across AC is equal to the voltage across the cell in the lower circuit, the ammeter can be connected without changing any voltage or current in the circuits. $\endgroup$ – Alfred Centauri May 12 '20 at 1:02
  • $\begingroup$ Also, note that comments are not for extended discussion. If, after you've thought about this some more, you still have questions, we should move this to a chat room. $\endgroup$ – Alfred Centauri May 12 '20 at 1:03
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The bulbs X Y and Z are not part of a normal potentiometer. Normally the battery E is balanced against AC and CB, but now AC is replaced by the combination of AC, X, Y, and Z. The internal resistance of the battery and meter don't matter, but these extra ones certainly do because they are in parallel with the battery instead of in series, and changing one of them will affect the balance point.

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  • $\begingroup$ thank you! however i am still unclear about something, if these bulbs are in parallel with the battery, musn't their potential drops match that of the battery in the secondary circuit? So overall wouldn't the potential drop still be the same and thus the balance length remains the same? $\endgroup$ – Lucas Tan May 11 '20 at 14:20
  • $\begingroup$ @LucasTan, I would recommend not accepting an answer less than two hours after posting your question. Generally speaking, it's best to give others here time to see and read your question and, if they wish, to volunteer their time to write and post an answer. Accepting the first answer right away tends to discourage others from making that effort. $\endgroup$ – Alfred Centauri May 11 '20 at 16:39
  • $\begingroup$ Could you elaborate on why would changing one of the bulbs affect the balance point? Thank you! Edit: Noted @AlfredCentauri, I will keep that in mind, thanks for pointing it out :) $\endgroup$ – Lucas Tan May 11 '20 at 22:56
  • $\begingroup$ @LucasTan, the potential drop across the bulb matches the drop across the battery, but the resistance of the globes has changed and so the current in the battery has changed. Because the battery has an internal resistance this results in an change to the voltage measured across the terminals of the battery, which will not be the same as the battery's ideal voltage (with no current flowing). $\endgroup$ – Peter May 11 '20 at 23:01
  • $\begingroup$ This seems to imply that the balance length depends on the terminal voltage of the battery, would I be correct to say that? If so, wouldn't adding more resistors in series also change the terminal voltage of the battery and thus imply that adding resistors in series would affect the balance length? However I know that adding resistors in series should not affect the balance length, hence I am confused about this terminal voltage concept. Thanks for the clarification! $\endgroup$ – Lucas Tan May 11 '20 at 23:13

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