Question

P.D. across a cell is 1.8 V when current of 0.5 A is drawn from it. The potential difference falls to 1.6V when current of 1A is drawn. Find emf of cell.

Answer 1

2.0 V

Answer 2

The relationship between the terminal potential difference ($V$) of a cell, its emf ($E$), the current drawn ($I$), and its internal resistance ($r$) is given by:

$V = E - Ir$

We are given two scenarios:

1. When $I = 0.5 \, \text{A}$, $V = 1.8 \, \text{V}$. Substituting these values into the formula, we get: $1.8 = E - 0.5r \quad (1)$

2. When $I = 1 \, \text{A}$, $V = 1.6 \, \text{V}$. Substituting these values into the formula, we get: $1.6 = E - 1r \quad (2)$

We now have a system of two linear equations with two unknowns, $E$ and $r$. We need to solve for $E$.

Subtract equation (2) from equation (1):

$(1.8) - (1.6) = (E - 0.5r) - (E - r)$ $0.2 = E - 0.5r - E + r$ $0.2 = 0.5r$ $r = \frac{0.2}{0.5} = \frac{2}{5} = 0.4 \, \Omega$

Now substitute the value of $r$ into either equation (1) or equation (2) to find $E$. Using equation (2):

$1.6 = E - 1(0.4)$ $1.6 = E - 0.4$ $E = 1.6 + 0.4$ $E = 2.0 \, \text{V}$

Thus, the emf of the cell is 2.0 V.