Question

A cell of emf $1.5V$ and internal resistance $0.5\Omega$ is connected to a conductor whose $V-I$ graph is an shown in the figure. Then, the terminal voltage of the cell is

• A. $0.75V$
• B. $1.0V$
• C. $1.25V$
• D. $1.5V$

Answer 1

The correct option is B $1.0V$

Given
The emf of the cell $=1.5V$
Internal resistance (r) $=0.5\Omega$
From graph we observe that
$V=I$
$\Rightarrow$ resistance of the conductor (R) $=\frac{V}{I}$ ($\because$ from ohm's law)
$=1\Omega$
so the total resistance of the current path $=R+r$
$=1+0.5\Omega$
$=1.5\Omega$

From ohm's law
the current in the circuit (i) $=\frac{\text{emf}}{\text{total resistance}}$

$=\frac{1.5}{1.5}A$

$=1A$
the voltage across the terminals $=$emf of cell$-$voltage drop across internal resistance
$=emf-(i\times r)$
$=1.5-(1\times 0.5)$
$=1.0V$