In the circuit shown in figure, if the diode forward voltage drop is 0.3 V, the voltage difference between A and B is

Options

A. 2.3 mA, 11.7 V
B. 2.34 A, 11.7 V
C. 2.43 mA, 1.17 V
D. 2.4 A, 12.3 V

Answer

A

Explanation

2.3 mA, 11.7 V
Forward bias for `Ge=0.3V`
`therefore` According to Kirchoff second law, for closed circuit,
`-0.3-5xx10^(3)I+12=0`
`therefore 11.7 =5xx10^(3)I therefore I=(11.7)/(5xx10^(3))`
`therefore I=2.34xx10^(-3)A`
`therefore I=2.34 mA`
and `V=IR`
`=2.34xx10^(-3)xx5xx10^(3)=11.7V`