From the circuit shown below, the maximum and minimum value of zener diode current are :

6 m A, 5 m A

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14 m A, 5 m A

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9 m A, 1 m A

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3 m A, 2 m A

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Solution

The correct option is C $9 m A, 1 m A$
The Zener diode acts as the voltage regulator in the circuit and the voltage supply is variable hence, excess of voltage over $50 V$ will be across $5 K Ω$ resistance.

Thus, minimum voltage across $5 K Ω$ resistance is $80 V - 50 V = 30 V$

Current through $5 K Ω$ resistance is $I_{1}$ (say)

$I_{1} = \frac{30}{5000} = 6 m A$

Also, maximum voltage across $5 K Ω$ resistance is $120 V - 50 V = 70 V$

Current through $5 K Ω$ resistance is $I_{1}^{'}$ ...(say)

$I_{1}^{'} = \frac{70}{5000} = 14 m A$
Now, voltage across load resistance of $10 K Ω$ is $50 V$

Thus, current through $10 K Ω$ will be
$I_{2} = \frac{50}{10000} = 5 m A$

Hence, maximum current through Zener diode is
$I_{z} = I_{1}^{'} - I_{2} = 14 - 5 = 9 m A$

Minimum current through Zener diode is
$I_{z} = I_{1} - I_{2} = 6 - 5 = 1 m A$

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I_{L}

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