Question

In the potentiometer circuit shown in figure , the internal resistance of the 12V battery is $1\Omega$ and length of the wire AB is 100 cm. when CB is 60 cm. the galvanometer show no deflection. the emf of cell E is (the resistance of wire AB is $3\Omega$)

• A. 3.6 V
• B. 2.4 V
• C. 3 V
• D. 4.5 V

Answer 1

The correct option is B 2.4 V

$Solution:$ We have been given a potentiometer circuit

diagram.In the given circuit diagram, the $12V$

battery has an internal resistance of $1\Omega$.Also, the

wire $AB$ is $100cm$ long. When the length of $CB$ is

$60cm$, the galvanometer shows no deflection.We are

also given that the resistance of wire $AB$ is $3\Omega$.Now,

we need to find out the E.M.F. of the cell E.The

current is equal to the voltage divided by the sum of

the external resistance, the internal resistance of the

battery and the resistance of wire $AB$. This can be

given as,

$I=\frac{12}{2+1+3}$

$\Rightarrow I=\frac{12}{6}$

$\Rightarrow I=2A$

Here, 12 is the voltage, 2 is the external resistance in

the circuit, 1 is the internal resistance of the battery,

and 3 is the resistance of wire $AB$.

Now, the E.M.F. of the cell $E$ is equal to the potential

difference across the section $AC$.

$E.M.F=I\times R$

$\Rightarrow E.M.F=2\times R$

For $100cm,R=3\Omega$.

As $AC=60cm,R=\frac{120}{100}\Omega$

Therefore, the E.M.F. of cell $E$ is equal to,

$E.M.F.=I\times R$

$\Rightarrow E.M.F.=2\times \frac{120}{100}$

$\therefore E.M.F.=2.4V$

Therefore, now we can conclude that the E.M.F. of

the cell $E$ is $2.4V$.

$So,the$ $correct$ $option:B$