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JEE Mains · Physics · STD 12 - Semicondutor electronics
- JEE Mains 2021
- Hard
- Numerical
In a given circuit diagram, a \(5 \,{V}\) zener diode along with a series resistance is connected across a \(50 \,V\) power supply. The minimum value of the resistance required, if the maximum zener current is \(90\, {mA}\) will be \(.....\,\Omega .\)

Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(500\)
Step-by-step Solution
Detailed explanation
Voltage across \({R}_{{L}}=5\, {V}\)
\(\Rightarrow {i}_{2}=\frac{5}{{R}_{{L}}}\)
Also voltage across \({R}=50-5=45\) \(volt\)
By \(v=i R \Rightarrow R=\frac{v}{i}=\frac{45}{i_{1}+i_{2}}\)
\({R}=\frac{45}{90 {m} {A}+\frac{5}{{R}_{{L}}}}\)
Current in zener diode is maximum when \({R}_{{L}} \rightarrow \infty\)
\(\left(i_{2} \rightarrow 0\right.\) and \(\left.i_{i}=i\right)\)
So \({R}=\frac{45}{90 {mA}}=500 \,\Omega\)
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