The barrier potential in a p-n junction diode is 0.3 volts . The current required is 6 mA. If a resistance of `200 Omega` is connected in series with the junction diode then the e.m.f. of the cell required for use in the circuit is

Options

A. 0.3 V
B. 1.2 V
C. 0.9 V
D. 1.5 V

Answer

D

Explanation

`V_(b)=0.3 V, I=6 mA , R= 200 Omega , E=?`
`I=(E-V_(b))/(R)`
`E=IR+V_(b)`
`=6xx10^(-3) xx200+0.3`
`=1.2+0.3`
`=1.5 V`