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

The circuit can be made as shown below.

`therefore " " V = E- ir`
where, r is internal resistance.
Also current `I = (E )/(R+r) rArr V = E-((E )/(R+r))r`
Putting the numerical values, we have
`E = 2V, r = 0.1 Omega, R = 3.9 Omega`
`V = 2-((2)/(3.9+0.1))xx0.1=1.95 V`