The cell has an emf of 2 V and the internal resistance of this cell is 0.1`Omega`, it is connected to resistance of 3.9`Omega`, the voltage across the cell will be

Options

A. 1.95V
B. 1.5V
C. 2V
D. 1.8V

Answer

A

Explanation

Key Idea When cell is giving current then the potential difference across its plates is less than its emf as potential drop across internal resistance occurs.
When a cell of emf E is connected to a resistance of `3.9Omega`, then the emf E of the cell remains constant, while voltage V goes on decreasing on taking more and more current from the cell.

`therefore V=E-ir`
Where, r is internal resitance.
Also, Current `i=(E)/(R+r)`
`therefore V=E-((E)/(R+r))r`
Putting the numberical values, we have
`E=2V, r=0.1Omega, R=3.9Omega`
`V=2-((2)/(3.9+0.1))xx0.1`
V=1.95volt