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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 \(.....\,\Omega .\)

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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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