Home
Class 12
PHYSICS
The emf of a cell is 2 V. When the termi...

The emf of a cell is 2 V. When the terminals of the cell is connected to a resistance `4Omega`. The potential difference across the terminals, if internal resistance of cell is `1Omega` is

A

`3/5` V

B

`8/5` V

C

`6/5` V

D

`5/8` V

Text Solution

Verified by Experts

The correct Answer is:
B
V=E-ir
Promotional Banner

Similar Questions

Explore conceptually related problems

According to this diagram , the potential difference across the terminals is ( internal resistance of cell =r)

The emf of a daniel cell is 1.08 V. When the terminals of the cells are connected to a ressitance of 3Omega , the potential difference across the terminals is found to be 0.6 V. Then the internal resistance of the cell is

A cell of emf 'E' is connected across a resistance R.The potential difference between the terminals of the cell is found to be "V" .The internal resistance of the cell must be:

The reading on a high resistance voltmeter. When a cell is connected across it is 2.2 V . When the terminals of the cell are connected to a resistance of 5 Omega the voltmeter reading drop to 1.8 V . Find the internal resistance of the cell.

The reading on a high resistance voltmeter when a cell is connected across it is 2.2 V. When the terminals of the cell are also connected to a resistance of 5 Omega , the voltmeter reading drops to 1.8 V. Find the internal resistance of the cell.