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

Answer

A

Explanation

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

`:. V = E -ir`
where `r` is current resistance.
Also current `i = (E)/(R + r)`
`:. 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 xx 0.1)) xx 0.1`
`V = 1.95 V`