Answer:
The answer is given in the explanation.
Explanation:
The circuit is as indicated in the attached figure.
From the analytical description the zener voltage is given as
[tex]V_z=V_z_o+I_zr_z[/tex]
Here
Vzo is the voltage at which the slope of 1/rz intersects the voltage axis it is equal to knee voltage.
The equivalent model is shown in the attached figure.
From the above equation, Vzo is calculated as
[tex]V_z_o=V_z-I_zr_z[/tex]
Here Vz is given as 7.5 V
Iz is given as 10 mA
rz is given as 30 Ω
Thus the Vzo is given as
[tex]V_z_o=V_z-I_zr_z\\V_z_o=7.4-30*10*10^{-3}\\V_z_o=7.5-0.3\\V_z_o=7.2 V[/tex]
The value of I_L is given as 5 mA
Now the expression of current is as
[tex]I=I_z+I_L\\I=10mA+5mA\\I=15 mA[/tex]
Now the resistance is calculated as
[tex]R=\dfrac{V-Vo}{I}\\R=\dfrac{10-7.2}{15*10^{-3}}\\R=186.66[/tex]
So the value of resistance is 186.66 Ω.
Considering the supply voltage is increased by 10%
V is 10-10%*10=10+1=11 so the
[tex]R=\dfrac{V-Vo}{I}\\186.66=\dfrac{11-V_o}{15*10^{-3}}\\V_o=8.2 V[/tex]
Considering the supply voltage is decreased by 10%
V is 10-10%*10=10-1=9 so the
[tex]R=\dfrac{V-Vo}{I}\\186.66=\dfrac{9-V_o}{15*10^{-3}}\\V_o=6.2 V[/tex]
Now if the supply voltage is 10% high and the value of Load is removed i.e I=Iz only which is 10mA
so
[tex]R=\dfrac{V-Vo}{I'}\\186.66=\dfrac{11-V_o}{10*10^{-3}}\\V_o=9.13 V[/tex]
Now the largest load current thus that the supply voltage is 10% low and the current of zener is knee current thus
[tex]V_z_o=V_z-I_zr_z\\V_z_o=7.5-30*0.5*10^{-3}\\V_z_o=7.5-0.015\\V_z_o=7.485 V[/tex]
[tex]R=\dfrac{V-Vo}{I'}\\186.66=\dfrac{9-7.485}{I}\\I=10.71 mA[/tex]
The load voltage is 7.485 V
