JEE Main 2025 Apr 2 Shift 1, Physics Q11: Semiconductors, p-n Junction Diode and Applications

JEE Main2025Apr 2, Shift 1Physics

A zener diode with 5V zener voltage is used to regulate an unregulated dc voltage input of 25 V. For a $400 Ω$ resistor connected in series, the zener current is found to be 4 times load current. The load current $(I_{L})$ and load resistance $(R_{L})$ are :

A
$I_{L} = 20$ mA; $R_{L} = 250 Ω$
B
$I_{L} = 10$ A; $R_{L} = 0.5 Ω$
C
$I_{L} = 0.02$ mA; $R_{L} = 250 Ω$
D
$I_{L} = 10$ mA; $R_{L} = 500 Ω$

Solution

Current through the series resistor. The voltage drop across the $400 Ω$ series resistor is $V_{in} - V_{Z} = 25 - 5 = 20 V$ . By KCL this resistor carries the sum of the zener current and the load current:

$I_{S} = I_{Z} + I_{L} = 4 I_{L} + I_{L} = 5 I_{L}$

$I_{S} = \frac{20 V}{400 Ω} = 0.05 A = 50 mA$

$\Rightarrow 5 I_{L} = 50 mA \Rightarrow I_{L} = 10 mA$

Load resistance. The voltage across the load equals the zener voltage:

$R_{L} = \frac{V _{Z}}{I _{L}} = \frac{5}{10 \times 1 0 ^{- 3}} = 500 Ω$

$∴$ Option (4) is correct.

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JEE Main 2025 Apr 2 Shift 1, Physics Q11: Semiconductors, p-n Junction Diode and Applications

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