Question

A cell of e.m.f. 1.5V and internal resistance $0.5\Omega $ is connected to a conductor whose V-I graph is as shown in figure. Then the terminal voltage of the cell is:

A) $0.75V$
B) $1.0V$
C) $1.25V$
D) $1.5V$

Answer 1

Hint: Recall that the resistance is defined as the opposition to the flow of electrons in a conductor. Also the Ohm’s law is used here. According to this law, the current flowing through any circuit is directly proportional to the voltage and inversely proportional to the resistance.

Complete step by step answer:
Step I:
Given that the e.m.f of the cell is $ = 1.5V$
And internal resistance$r = 0.5\Omega $
The resistance of the conductor can be known by using Ohm’s law. As per Ohm’s law
$V = IR$
$\Rightarrow R = \dfrac{V}{I}$
Step II:
But since the graph is linear, this means that
$V = I$
Therefore, $R = 1\Omega $
The total resistance of the conductor is given by= Resistance of the conductor + internal resistance
$\Rightarrow R_{total} = 1 + 0.5$
$\Rightarrow R_{total}= 1.5\Omega $
Step III:
Again using Ohm’s law to find the value of current in the circuit.
$I = \dfrac{V}{R}$
e.m.f. of a cell is the potential difference across the terminals of a cell when no current is flowing in the circuit. So finding the value of current
$\Rightarrow I = \dfrac{{1.5}}{{1.5}}$
$\Rightarrow I = 1A$

The potential difference that is applied between the terminals of the battery is known as terminal voltage. The voltage across the terminal is given by
=e.m.f.-voltage drop across internal resistance
$\Rightarrow $ Terminal voltage $ = 1.5 - (1 \times 0.5)$
$\Rightarrow $ Terminal voltage $ = 1.0V$

Therefore the terminal voltage across the circuit is $1V$. Hence Option B is the right answer.

Note:
It is to be noted that the terminal voltage is always less than the e.m.f. of the cell. This is because when the current passes through the internal resistance of the conductor or the circuit then it causes some potential drop. But when no current passes the voltage across the terminals is the e.m.f.