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
![V_z=V_z_o+I_zr_z](https://tex.z-dn.net/?f=V_z%3DV_z_o%2BI_zr_z)
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
![V_z_o=V_z-I_zr_z](https://tex.z-dn.net/?f=V_z_o%3DV_z-I_zr_z)
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
![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](https://tex.z-dn.net/?f=V_z_o%3DV_z-I_zr_z%5C%5CV_z_o%3D7.4-30%2A10%2A10%5E%7B-3%7D%5C%5CV_z_o%3D7.5-0.3%5C%5CV_z_o%3D7.2%20V)
The value of I_L is given as 5 mA
Now the expression of current is as
![I=I_z+I_L\\I=10mA+5mA\\I=15 mA](https://tex.z-dn.net/?f=I%3DI_z%2BI_L%5C%5CI%3D10mA%2B5mA%5C%5CI%3D15%20mA)
Now the resistance is calculated as
![R=\dfrac{V-Vo}{I}\\R=\dfrac{10-7.2}{15*10^{-3}}\\R=186.66](https://tex.z-dn.net/?f=R%3D%5Cdfrac%7BV-Vo%7D%7BI%7D%5C%5CR%3D%5Cdfrac%7B10-7.2%7D%7B15%2A10%5E%7B-3%7D%7D%5C%5CR%3D186.66)
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
![R=\dfrac{V-Vo}{I}\\186.66=\dfrac{11-V_o}{15*10^{-3}}\\V_o=8.2 V](https://tex.z-dn.net/?f=R%3D%5Cdfrac%7BV-Vo%7D%7BI%7D%5C%5C186.66%3D%5Cdfrac%7B11-V_o%7D%7B15%2A10%5E%7B-3%7D%7D%5C%5CV_o%3D8.2%20V)
Considering the supply voltage is decreased by 10%
V is 10-10%*10=10-1=9 so the
![R=\dfrac{V-Vo}{I}\\186.66=\dfrac{9-V_o}{15*10^{-3}}\\V_o=6.2 V](https://tex.z-dn.net/?f=R%3D%5Cdfrac%7BV-Vo%7D%7BI%7D%5C%5C186.66%3D%5Cdfrac%7B9-V_o%7D%7B15%2A10%5E%7B-3%7D%7D%5C%5CV_o%3D6.2%20V)
Now if the supply voltage is 10% high and the value of Load is removed i.e I=Iz only which is 10mA
so
![R=\dfrac{V-Vo}{I'}\\186.66=\dfrac{11-V_o}{10*10^{-3}}\\V_o=9.13 V](https://tex.z-dn.net/?f=R%3D%5Cdfrac%7BV-Vo%7D%7BI%27%7D%5C%5C186.66%3D%5Cdfrac%7B11-V_o%7D%7B10%2A10%5E%7B-3%7D%7D%5C%5CV_o%3D9.13%20V)
Now the largest load current thus that the supply voltage is 10% low and the current of zener is knee current thus
![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](https://tex.z-dn.net/?f=V_z_o%3DV_z-I_zr_z%5C%5CV_z_o%3D7.5-30%2A0.5%2A10%5E%7B-3%7D%5C%5CV_z_o%3D7.5-0.015%5C%5CV_z_o%3D7.485%20V)
![R=\dfrac{V-Vo}{I'}\\186.66=\dfrac{9-7.485}{I}\\I=10.71 mA](https://tex.z-dn.net/?f=R%3D%5Cdfrac%7BV-Vo%7D%7BI%27%7D%5C%5C186.66%3D%5Cdfrac%7B9-7.485%7D%7BI%7D%5C%5CI%3D10.71%20mA)
The load voltage is 7.485 V