Consider a Zener diode connected in reverse bias with a 15 V battery and a `10 k omega` resistor in series. If the breakdown voltage of the diode is 4.5 V, then find the current flowing through the Zener in mA:

Answer

`01.05`

Explanation

`I=(V_"in"-V_z)/R_s`
`=(15-4.5)/(10 xx 10^3)`
`=(10.5/10^4)`
`=1.05 xx 10^(-3)A`
`:. I=1.05 mA`