A diode made of silicon has a barrier potential of `0.7 V` and a current of `20 mA` passes through the diode when a battery of `emf` ` 3 V` and a resistor is connected to it. The wattage of the resistor and diode are

Options

A. 46 W
B. 45 W
C. 14 W
D. 14 mW

Answer

B

Explanation

`V_(b) V, I=20 mA, E=3 V, P =? `
`I=(E-V_(b))/(R)`
`R=(E-V_(b))/(I)`
`=(3-0.7)/(20xx10^(-3))=(2.3xx10^(3))/(20)=115 Omega`
Now `=I^(2)R=400xx10^(-6)xx15`
`=46000xx10^(-6)`
`46 10^(-3)` W
=46 mW