In the circuit given, the current through Zener diode is

Options

A. 10 mA
B. 6.67 mA
C. 3.33 mA
D. 5 mA

Answer

C

Explanation

According to the question, the circuit is

Since, potential drop across a Zener diode is constant. So, applying KVL in the circuit.
`" "15=IR_(1)+V_(z)`
`rArr " "15=I(500)+10" "(because V_(z)=10V)`
`rArr " "I=(15-10)/500=1 times 10^(-2)A=10mA`
There is potential drop of 10 V across the `1500 Omega` resistor.
So, `" "I_(1)=V/R_(2)=10/1500`
`" "=6.66mA`
Therefore, the current through the zener diode,
`" "I_(Rs.)=I-I_(1)=10-6.66=3.33 mA`
`therefore` The correct option is (c).