JEE Mains · Physics · STD 12 - 14. Semicondutor electronics
A zener diode having zener voltage \(8\, {V}\) and power dissipation rating of \(0.5\, {W}\) is connected across a potential divide arranged with maximum potential drop across zener diode is as shown in the diagram. The value of protective resistance \({R}_{{p}}\) is \(....\,\Omega\)
- A \(123\)
- B \(456\)
- C \(192\)
- D \(219\)
Answer & Solution
Correct Answer
(C) \(192\)
Step-by-step Solution
Detailed explanation
\(P=V i\) \(0.5=8 i\) \(i=\frac{1}{16}\, A\) \(E=20=8+i R_{p}\) \(R_{P}=12 \times 16=192\, \Omega\)
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