Physics Q45 – JEE-MAIN 2025

Home Physics SemiconductorsQuestion #470

A zener diode with

5 V

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 :

Clue 1 of 3: Zener Diode Regulator

Understand the voltage and current distribution in a zener diode voltage regulator circuit.

🥷

Ninja Strategy•Calculate Total Current First

First, calculate the total current through the series resistor using the voltage drop across it and its resistance, then use the given ratio of zener to load current to find the load current.

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● Video Walkthrough

Step 1: Calculate total current through series resistor✦ Active

The voltage across the series resistor $R_{S}$ is the difference between the input voltage and the zener voltage (which is the regulated output voltage):

V_{S} = V_{i n} - V_{Z} = 25 V - 5 V = 20 V

The total current $I_{S}$ flowing through $R_{S}$ is then calculated using Ohm's Law:

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

Step 2: Determine load current (

I_{L}

)○ Expand

The total current $I_{S}$ splits into the zener current $I_{Z}$ and the load current $I_{L}$ . So, $I_{S} = I_{Z} + I_{L}$ . We are given that the zener current is 4 times the load current, i.e., $I_{Z} = 4 I_{L}$ . Substituting this into the current equation:

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

Now, we can find the load current $I_{L}$ :

I_{L} = \frac{I_{S}}{5} = \frac{50 mA}{5} = 10 mA

Step 3: Calculate load resistance (

R_{L}

)○ Expand

The voltage across the load resistor $R_{L}$ is equal to the zener voltage, $V_{L} = V_{Z} = 5 V$ . Using Ohm's Law for the load:

R_{L} = \frac{V_{L}}{I_{L}} = \frac{5 V}{10 mA} = \frac{5 V}{10 \times 10^{- 3} A} = 500 Ω

💡 Teacher's Secret Hint

Ensure consistent units (Volts, Amperes, Ohms) for calculations.

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Physics Question 45 – JEE-MAIN 2025

Clue 1 of 3: Zener Diode Regulator

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