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

Answer

C

Explanation

(c ) The voltage across cell terminal will be given by `V = E - ir`
Where `r` is internal resistance.
Also current `i= (E)/(R + r)`
`:. V = E - ((E)/(R + r)) r`
Putting the numerical values, we have
`E == 2 V, R = 0.1 Omega, R = 3.9 Omega`
`V = 2 - ((2)/(3.9 + 0.1)) xx 0.1`
`V = 1.95 V`