0
$\begingroup$

I could not understand why the emf of the main cell (labelled E) in the circuit (can be a battery eliminator) should be greater than the emf of the cells (E1 and E2) used in the secondary circuit. circuit

Please help.

$\endgroup$
0
$\begingroup$

The maximum possible potential difference across the potentiometer wire $PQ$ is $\mathcal E$.
In practice it would be less because of the drop in potential across the ammeter, the cell $B$ due to its internal resistance and the rheostat.

At balance, no current through the galvanometer $G$, the potential drop across the wire $PJ$ must equal the emf, $\mathcal E_1$ or $\mathcal E_2$, of the cell connected to the circuit.

If the potential difference across the whole wire $PQ$ is less than $\mathcal E_1$ or $\mathcal E_2$ then no such balance point can be found.
So the emf of cell $B$ must be bigger than $\mathcal E_1$ (or $\mathcal E_2$) otherwise there will be no balance point.

$\endgroup$
0
$\begingroup$

Welcome to Physics SE. You should give a little more detail about your circuit and what you are trying to achieve using it.

So the purpose of this experiment to find the null point that is when there is no current passing through the Galvanometer and compare the emf of the two cells.

If $\mathcal E_1$ and $\mathcal E_2$ are bigger than $\mathcal E$, then a current will always flow through the galvanometer and so nu null point will ever occur.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.