Correct Answer - D
The voltage drop across `R_2` is
`V_(R_2)=V_Z=10 V`
The current through, `R_2` is `I_(R_2)=(V_(R_2))/R_2=(10V)/(1500 Omega)=0.667xx10^(-2) A`
`=6.67xx10^(-3)` A = 6.67 mA
The voltage drop across `R_1` is
`V_(R_1)=15 V-V_(R_2)=15 V-10 V=5 V `
The current through `R_1` is
`I_(R_1)=V_(R_1)/R_1=(5V)/(500Omega)=10^(-2)A=10xx10^(-3)A =10 mA`
The current through the zener diode is
`I_Z=I_(R_1)=I_(R_2)=(10-6.67)mA =3.33mA`
