A cell is a source of electric current in the electrical circuit. The Potential Difference between terminals of a cell in an open circuit (when no current is drawn) is called electromotive force (emf) of the cell. When electrical circuit is closed and current is drawn from the terminal Potential Difference between two terminals is called terminal potential difference (v) of the cell. The cells can be connected in series as well as in parallel combinations. Like resistor cell also offers opposition to the flow of current. This opposition offered by cell is called internal resistance of the cell.

When a cell is connected across external resistance 5 Ω, a current of 0.25 A flows through the circuit. If the external resistance is replaced by 2 Ω, a current of 0.5 A flows through it. The emf of the cell in the circuit is:

1.5 V

The correct answer is Option (4) → 1.5 V

We can solve this problem using the concept of the internal resistance (r) of the cell and the relation between the emf (E), terminal potential difference (v), and the total resistance of the circuit.

$E= I(R+r)$

where,

E = emf of the cell

I = Current in the circuit

R = External Resistance

r = Internal Resistance of the cell

CASE 1:

$R_1=5Ω$

$I_1​=0.25A$

from the formula:

$E=I_1(R_1+r)$

$E=0.25(5+r)$  ...(1)

CASE 2:

$R_2=2Ω$

$I_2=0.5A$

from the formula:

$E=I_2(R_2+r)$

$E=0.5(2+r)$  ...(2)

from eq. (1) and eq. (2), we equate the expression for E.

$0.25(5+r)=0.5(2+r)$

$1.25++0.25r=1+0.5r$

$0.25=0.25r$

$r=1Ω$

By substituting $r=1Ω$ in equation (1)

$E=0.25(5+1)$

$=0.25×6=1.5V$