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33 - Comparing the EMF of two cells using Potentiometer.md

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  • Setup

The theory is if the galvanometer reading is 0, the voltage drop in the potentiometer circuit is equal to the the potential drop in the external circuit.

Here we have connected the $E_1$ cell using the 2 way switch and the potential difference of $E_1$ cell in this case is proportional to the $l_1$ length. We can do the same for the other cell and compare them. Using that we can build up 2 equations.

$$ \begin{align*} k \to& Potential\ Gradient\\\ E_1 \propto l_1\\ E_1 &= kl_1\\ E_2 \propto l_2\\ E_2 &= kl_2\\\ \therefore \frac{E_1}{E_2} &= \frac{l_1}{l_2} \end{align*} $$

If we want to go for a graph we can do that as follows. $$ \begin{align*} \frac{E_1}{E_2} &= \frac{l_1}{l_2}\ \ l_1 &= (\frac{E_1}{E_2})l_2 \ y &= mx \end{align*} $$

Here, the way we change the lengths $l_1$ and $l_2$ for multiple readings is by changing the potential gradient $k$. This can be done by changing the resistance in the potentiometer circuit. For that we can use the rheostat

Important Points

  • If the balance lengths are too small or big, and needed to make bigger, what is the adjustment we can do?
Change the resistance using the rhoestat (in the potentiometer circuit) accordingly
  • What the main advantage of using the potentiometer over the voltmeter?
Practically with the voltmeter  some current will be drawn while measuring the potential difference and the reading will deviate a little (V = E - Ir). But with potentiometer no current will be drawn so the measurement we get will be much more accurate.
  • What is the main characteristic of the accumulator? (Cell used in the potentiometer circuit?
It should be able to provide a constant potential difference for a longer time without running down.
  • What is the purpose of the variable resistor in series with the accumulator in the circuit?
To change the potential gradient and take multiple readings for l1 nad l2
  • What would be the reasons if you can't find the balanced length using the external circuit?
1. There maybe a loose connection.
2. Positive terminal of the cells and the accumilator (cell in the potentiometer circuit) may not be connected to the same end of the potentio cable.

(this is how it should be)

3. EMF of the extrernal cells maybe greater than the acuumulator (potentiometer circuit cell)
  • What are the ways we can correct the circuit if the emf of external cells are greater than potentiometer circuits'?
- Applying 2 accumilators for the potentiometer circuit in series.
- Replacing the accumilator with a one having a greater EMF.
  • Out of these 2, what is the preferred method and why?
Using an accumilator with a bigger EMF.

Because if we use 2 cells in series not only the voltage but also the internal resistance will increase causing issues with our calculations.