JEE Physics 2026 Question with Solution

Given: Zener breakdown voltage $V_Z = 10\,\text{V}$, maximum power dissipation $P_{\max} = 0.4\,\text{W}$, and input voltage $V_{\text{in}} = 15\,\text{V}$.

Find: The protective resistance $R$.

From the maximum power rating of the Zener diode,

So,

The voltage across the protective resistance is,

Now calculate the resistance using Ohm's law,

The solution also states $R \approx 5\,\Omega$, but this contradicts the shown calculation. The worked steps clearly give $125\,\Omega$, so that is the correct numerical answer.

Therefore, the protective resistance is $125\,\Omega$.

Given: The diode must not exceed its maximum power rating.

Find: Which numerical value is consistent with the working.

If the resistor were $5\,\Omega$, then the current would be

The Zener power would then be

which is far greater than the allowed $0.4\,\text{W}$.

Hence $5\,\Omega$ cannot be correct. The only value consistent with the maximum power condition is $125\,\Omega$.