The correct answer is 2

Minimum value of RL for which the diode is shorted is
\(\frac{R_L}{R_L}\)+100×10=8 ⇒RL=400 Ω
For maximum value of RL, current through diode is 10 mA.
So
\(i_R= i_{R_L} +I_{ZM}\)
\(\frac{2}{100}\)=\(\frac{8}{R_L}\)+10×10−3
10×10−3=\(\frac{8}{R_L}\)
RL = 800 Ω
So
\(\frac{R_{Lmax}}{R_{Lmin}}=\frac{800}{400}=2\)
So, ratio of maximum and minimum value of RL is 2
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