Physics Q46 – JEE-MAIN 2026

Home Physics SemiconductorsQuestion #463

A diode has Zener voltage of

10 V

and maximum power dissipation of

0.5 W

, then the minimum resistance to be used in series with this diode for safety when it is connected to a

25 V

power supply is _______

Ω

Enter Calculated Numerical Value:

Clue 1 of 3: Zener Diode Characteristics

Understand that a Zener diode maintains a constant voltage across itself when operating in its breakdown region.

🧠 Full Solution Path

Step 1: Calculate Maximum Zener Current✦ Active

The maximum power dissipation ( $P_{m a x}$ ) of the Zener diode is given as $0.5 W$ , and its Zener voltage ( $V_{Z}$ ) is $10 V$ . The maximum current ( $I_{Z, m a x}$ ) the Zener diode can safely handle is calculated using the power formula:

P_{m a x} = V_{Z} \times I_{Z, m a x}

0.5 W = 10 V \times I_{Z, m a x}

I_{Z, m a x} = \frac{0.5 W}{10 V} = 0.05 A

Step 2: Calculate Voltage Across Series Resistor○ Expand

The Zener diode is connected in series with a resistor to a $25 V$ power supply ( $V_{S}$ ). Since the Zener diode maintains a constant voltage of $V_{Z} = 10 V$ across itself, the voltage drop across the series resistor ( $V_{R}$ ) is:

V_{R} = V_{S} - V_{Z}

V_{R} = 25 V - 10 V = 15 V

Step 3: Calculate Minimum Series Resistance○ Expand

For safety, the current flowing through the series resistor must not exceed the maximum safe Zener current, $I_{Z, m a x}$ . Using Ohm's law, the minimum series resistance ( $R_{m i n}$ ) required is:

R_{m i n} = \frac{V_{R}}{I_{Z, m a x}}

R_{m i n} = \frac{15 V}{0.05 A} = 300 Ω

💡 Teacher's Secret Hint

Ensure the current through the series resistor does not exceed the maximum Zener current to prevent damage to the diode.

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Physics Question 46 – JEE-MAIN 2026

Clue 1 of 3: Zener Diode Characteristics

🧠 Full Solution Path