Question

A cell supplies a current of $0.9\;A$ through a $2\Omega $ resistor and a current of $0.3\;A$ through a $7\Omega $ resistor. The internal resistance of the cell is:-
(A) $1.0\Omega $
(B) $0.5\Omega $
(C) $2.0\Omega $
(D) $1.2\Omega $

Answer 1

Hint: Since in both cases the cell is the same, we can assume that the e.m.f. and the internal resistance in both cases is the same. The internal resistance of a cell is always in series with the other resistance. Therefore, by equating the net e.m.f. in both cases, we can evaluate the value of internal resistance as it is similar to a set of linear equations with two variables.
Formula used:
$E = I(R + r)$

Complete step by step answer:
It is given in the question, that in the first case,
The current passing through the resistor is, ${I_1} = 0.9\;A$
The resistor in the circuit, ${R_1} = 2\Omega $
Let $E$ be the e.m.f. across the cell and $r$ be the internal resistance of this cell. Internal resistance is always assumed to be in series with the other resistors ( or other elements) of a circuit.
Then the e.m.f in this circuit is given by-
$E = {I_1}({R_1} + r)$
Putting the values,
$E = 0.9(2 + r)$ …$(1)$
In the second case,
The current passing through the resistor is, ${I_2} = 0.3\;A$
The resistor in the circuit, ${R_2} = 7\Omega $
Then the e.m.f. in the circuit is given by-
$E = {I_2}({R_2} + r)$
Putting the values,
$E = 0.3(7 + r)$ …$(2)$
We can equate both equations $(1)$ and $(2)$ because e.m.f is the same in both of the cases.
$0.9(2 + r) = 0.3(7 + r)$
$ \Rightarrow 3(2 + r) = 7 + r$
$ \Rightarrow 6 + 3r = 7 + r$
Moving similar terms on the same side,
$3r - r = 7 - 6$
$2r = 1$
Which gives,
$r = 0.5\Omega $

Therefore, the internal resistance of the cell is $r = 0.5\Omega $, and option (B) is the correct answer.

Note: Internal resistance is caused due to the type of electrolyte used for storing the charge. It is present inside the structure of the cell but it can be considered to be a separate resistance that is connected in series with the circuit. Therefore, it can be directly added to any other resistor present in the circuit.