The potential gradients on the potentiometer wire are $V_{1}$ and $V_{2}$ with an ideal cell and a real cell of same emf in the primary circuit then

V_{1} = V_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

V_{1} > V_{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

V_{1} < V_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

V_{1} \leq V_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $V_{1} > V_{2}$
In an ideal cell the internal resistance is zero.
$∴$ Whole P.D is across potentiometer.
But in an real cell the internal resistance is not zero.
$\Rightarrow$ There will be some potential drop across internal resistance.
$∴$ P.D across potentiometer < Emf of cell.
$∴ V_{1} > V_{2}$ .

Suggest Corrections

Similar questions

$V_{1}, V_{2} a n d V_{3}$ are the potential differences across the $1 Ω, 2 Ω a n d 3 Ω$ resistors and the flow of current is 5 A in a circuit .
Which of the following shows the correct values of $V_{1}, V_{2} a n d V_{3}$ measured in volts?

A spherical shell of radius 10 cm is carrying a charge q. if the electric potential at distances 5 cm, 10 cm and 15 cm from the centre of the spherical shell is $V_{1}, V_{2} a n d V_{3}$ respectively, then

In the circuit shown in the figure, reading of voltmeter is $V_{1}$ when only $S_{1}$ is closed, $V_{2}$ when only $S_{2}$ is closed and $V_{3}$ when both $S_{1}$ and $S_{2}$ are closed. Choose the correct relation between $V_{1}$ , $V_{2}$ and $V_{3}$ .