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Step-by-Step Solution
Step 1: Identify the given data
• Zener breakdown voltage, V_z = 8\,\text{V}
• Maximum zener current, I_{ZM} = 10\,\text{mA}
• Input supply voltage, V_i = 10\,\text{V}
• Series resistor, R = 100\,\Omega
• Variable load resistor, R_L
• We need the ratio R_{L\,\max}/ R_{L\,\min} .
Step 2: Determine R_{L\,\min}
For the Zener diode to just enter breakdown, it must have 8\,\text{V} across it. Hence, the potential division between R_L and R should place 8\,\text{V} across R_L . Since the total supply is 10\,\text{V} , the fraction of 10\,\text{V} across R_L is:
\frac{R_L}{R_L + R} \times 10 = 8.
Substituting R = 100\,\Omega , we get:
\frac{R_L}{R_L + 100} \times 10 = 8
\quad \Longrightarrow \quad
8 \bigl(R_L + 100\bigr) = 10\,R_L
\quad \Longrightarrow \quad
8R_L + 800 = 10R_L
\quad \Longrightarrow \quad
2R_L = 800
\quad \Longrightarrow \quad
R_L = 400\,\Omega.
Thus,
R_{L\,\min} = 400\,\Omega.
Step 3: Determine R_{L\,\max}
The maximum value of R_L occurs when the Zener diode carries its maximum current I_{ZM} = 10\,\text{mA} . In Zener breakdown, the voltage across the diode is 8\,\text{V} , so the voltage drop across the series resistor R is:
V_R = V_i - V_z = 10\,\text{V} - 8\,\text{V} = 2\,\text{V}.
Therefore, the total current through the series resistor R is:
i_R = \frac{V_R}{R} = \frac{2\,\text{V}}{100\,\Omega} = 0.02\,\text{A} = 20\,\text{mA}.
This current divides into two parts: one through the load resistor ( i_{R_L} ) and one through the Zener diode ( I_{ZM} ). Hence,
i_R = i_{R_L} + I_{ZM}.
Substituting i_R = 20\,\text{mA} and I_{ZM} = 10\,\text{mA} ,
i_{R_L} = 20\,\text{mA} - 10\,\text{mA} = 10\,\text{mA}.
Since i_{R_L} = \frac{V_L}{R_L} and the voltage V_L across R_L is 8\,\text{V} (in Zener breakdown region), we have:
i_{R_L} = \frac{8\,\text{V}}{R_L} = 10\,\text{mA}.
Therefore,
R_L = \frac{8\,\text{V}}{10 \times 10^{-3}\,\text{A}} = 800\,\Omega.
Thus,
R_{L\,\max} = 800\,\Omega.
Step 4: Calculate the ratio
Finally, the required ratio is:
\frac{R_{L\,\max}}{R_{L\,\min}}
\;=\;
\frac{800\,\Omega}{400\,\Omega}
\;=\;
2.
Answer
The ratio of maximum to minimum value of R_L is 2 .
