Question

When a $2 \Omega $ resistance is connected across the terminal of a cell, the current is $0.5A$. When the resistance is increased to $5 \Omega$, the current becomes $0.25A$. The e.m.f. of cell is-
A) $1.0 V$
B) $1.5 V$
C) $2.0 V$
D) $2.5 V$

Answer 1

Hint
There are two resistance connected circuits, one is given resistance and another internal resistance of the cell. We apply ohm’s law on both conditions and find the e.m.f. of cell. We equate the current in both cases and get some equation in terms of e.m.f. and internal resistance.

Complete step by step solution
Let $E$ be the emf of call and $R$ is internal resistance of the cell.
For the first case, external resistance is $2 \Omega$ and the current in the circuit is $0.5 A$. Then-
$\dfrac{E}{{R + 2}} = 0.5$ or $E = 0.5(R + 2)$ -(1)
Similarly, in the second case resistance is $5 \Omega$and current in circuit is $0.25 A$. Then
$E = 0.25(R + 5)$ -(2)
From equation (1) and (2), we get
$0.5(R + 2) = 0.25(R + 5)$
$R = 1 \Omega$
Put value of R in equation (1), and we get
$E = 0.5(2 + 1) = 1.5V$.
Hence, the correct answer is option (B).

Note
Internal resistance of a cell is resistance offered by chemical and material inside the cell. If we have given potential differences between terminals of a cell, then we don’t consider internal resistance because internal resistance is already taken. If we want to use more percentage of power of the cell then external resistance must be greater than internal resistance.