Question

If the internal resistance of a cell is $20\Omega $ and resistance of the circuit is also $20\Omega $. Will the current remain the same whether the given $n$ identical cells are in series or in parallel?

Answer 1

Hint: In the given question, we are provided with the internal resistance of the cell and the resistance of the circuit and we have to compare the current when the cells are connected in series or in parallel. So, we find out the current in both the cases using the Ohm’s Law as \[V = IR\] by substituting the smf of the cell and the resistance in the circuit including the internal resistance of the cell. Then, we compare the currents in both the cases to get the final answer.
Formulae Used:
If n cells of e.m.f. ${V_1}$, ${V_2}$, ${V_3}$ , … are connected in series, then the effective e.m.f. is \[{V_1} + {V_2} + {V_3} + ... + {V_n}\]
If n cells of e.m.f. ${V_1}$, ${V_2}$, ${V_3}$ , … are connected in parallel, then the effective e.m.f. is \[\dfrac{{{V_1} + {V_2} + {V_3} + ... + {V_n}}}{n}\]

Complete answer:
It is given that there are $n$ identical cells and each has internal resistance of $20\Omega $ The resistance of the circuit is also $20\Omega $.
It is required to verify whether the current is the same or not for both series and parallel connection of the given $n$ identical cells.
Let e.m.f. of each cell is $V$.
Case I-
Consider the given $n$ cells are connected in series.
Then effective e.m.f. $ = nV$
Since cells are connected in series, the internal resistances are also in series connection.
The net internal resistance is ${r_{eff}} = 20n$.
The resistance in the circuit is $R = 20\Omega $
The current flowing through the circuit is ${I_S} = \dfrac{{{\text{effective emf}}}}{{{r_{eff}} + R}}$
Or ${I_S} = \dfrac{{nV}}{{20n + 20}}$
Case II-
Now consider the cells are connected in parallel.
The effective e.m.f. of the circuit is $V$
Since cells are connected in parallel, the internal resistances are also in parallel connection.
The net internal resistance is ${r_{eff}} = \dfrac{{20}}{n}$
The current flowing through the circuit is ${I_P} = \dfrac{{{\text{effective emf}}}}{{{r_{eff}} + R}}$
Or ${I_P} = \dfrac{V}{{\dfrac{r}{n} + R}}$
Further simplifying
$ \Rightarrow {I_P} = \dfrac{{nV}}{{r + nR}}$
Substitute the values of $r$ and $R$ in the above formula
$ \Rightarrow {I_P} = \dfrac{{nV}}{{20 + 20n}}$
Now compare the current obtained in both the cases.
$ \Rightarrow {I_S} = {I_P}$
Hence, the current is the same for both the cases.

Note:
Since it is not given how the positive and negative terminals are connected, we assume that in series the positive terminal of a cell is connected to the negative terminal of the neighbour cell. And in parallel connection, all the positive terminals are connected in one junction and all the negative terminals connected in another junction.