Question

When $0.5A$ current is drawn from a cell, its potential difference is $1.8V$which reduces to $1.6V$ on drawing a current of $1.0A$. The e.m.f. of the cell is:

• A. $2V$
• B. $4V$
• C. $6V$
• D. $8V$

Answer 1

The correct option is A $2V$

In the first circuit,

$i_1=0.5Amp$ $V_{AB}=1.8V$

$V_{AB}=E-i_{1}r$

$1.8=E-\frac{r}{2}----\left(1\right)$

Now, in the second circuit,

$i_2=1Amp$ $V_{AB}=1.6V$

$V_{AB}=E-i_{2}r$

$1.8=E-r----\left(2\right)$

$by\left(2\times equation\left(1\right)\right)-\left(equation\left(2\right)\right)weget,$

$2=E$

$E=2volt$

and,

$E.m.f=2volts$

Hence, the option $\left(A\right)$ is the correct answer.