JEE Mains · Physics · STD 12 - Semicondutor electronics
- JEE Mains 2026
- Easy
- Numerical
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 ______ \(\Omega\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) 300
Step-by-step Solution
Detailed explanation
Given Zener voltage \(V_Z = 10\) V and maximum power dissipation \(P_{max} = 0.5\) W.
The maximum safe current through the Zener diode is given by \(I_{max} = \dfrac{P_{max}}{V_Z} = \dfrac{0.5}{10} = 0.05\) A.
The voltage drop across the series resistor is \(V_R = V_S - V_Z = 25 - 10 = 15\) V.
For safety, the current through the circuit should not exceed \(I_{max}\). Thus, the minimum resistance \(R\) is \(R = \dfrac{V_R}{I_{max}} = \dfrac{15}{0.05} = 300\) \(\Omega\).
Answer: \(300\)
The maximum safe current through the Zener diode is given by \(I_{max} = \dfrac{P_{max}}{V_Z} = \dfrac{0.5}{10} = 0.05\) A.
The voltage drop across the series resistor is \(V_R = V_S - V_Z = 25 - 10 = 15\) V.
For safety, the current through the circuit should not exceed \(I_{max}\). Thus, the minimum resistance \(R\) is \(R = \dfrac{V_R}{I_{max}} = \dfrac{15}{0.05} = 300\) \(\Omega\).
Answer: \(300\)
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