Question

A zener diode of voltage `V_(Z)(=6` volt) is used to maintain a constant voltage across a load resistance `R_(L)(=1000Omega)` by using a series resistance `R_(S)(+100Omega)`. If the emf or source is `E(=9V)`, clculate the value of current through series resistance, zener diode and load resistance. What is the power being dissipated in zener diode.

Answer 1

Here, `E=9V,V_(Z),R_(L)=1000Omega` and `R_(S)=100Omega`
Potential drop across series resistor `V=E-V_(Z)=9-6=3V`
Current through series resistance `R_(S)` is `I=(V)/(R)=(3)/(100)=0.03A`
Current through load resistance `R_(L)` is `I_(L)=(V_(Z))/(R_(L))=(6)/(1000)=0.006A`
Current through zener diode is `I_(Z)=I-I_(L)=0.03-0.006=0.024A`
Power dissipated in zener diode is `P_(Z)=V_(Z)I_(Z)=6xx0.024=0.144`watt