Question

The e.m.f. of the cell is

• A. $1.2V$
• B. $4.8V$
• C. $1.8V$
• D. $2V$

Answer 1

Answer: (C)

$1.8V$

In the first case, there are 2 resistors of 2 ohms each connected in parallel.

The combined resistance is given as $\frac{1}{2}+\frac{1}{2}=\frac{1}{1}=1ohms$.
In the second case, there are 2 resistors of 2 ohms each connected in series. The combined resistance is given as $2+2=4$ $ohms$.

The current in the circuit is given by the formula $I=\frac{\epsilon }{R+r}$ where,
$\epsilon$- emf of the cell
R- external resistance
r- internal resistance of the cell.

The current in the 1 st case is given as $1.2=\frac{\epsilon }{1+r},thatis,1.2+1.2r=\epsilon$ - eqn - 1
The current in the 2nd case is given as $0.4=\frac{\epsilon }{4+r},thatis,1.6+0.4r=\epsilon$ - eqn - 2

Solving eqn 1 and eqn 2 for r and $\epsilon$, we get, $r=0.5$ ohms and $\epsilon$ $=1.8$ volts

Hence, the emf of the cell is given as 1.8 volts.