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JEE Mains · Physics · STD 12 - Semicondutor electronics
  • JEE Mains 2021
  • Medium
  • Numerical

The value of power dissipated across the zener diode \(\left( V _{ z }=15 V \right)\) connected in the circuit as shown in the figure is \(x \times 10^{-1}\) watt.  The value of \(x ,\) to the nearest integer, is ..............

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Answer & Solution

Correct Answer

(A) \(5\)

Step-by-step Solution

Detailed explanation

Voltage across \(R _{ S }=22-15=7 V\)

Current through \(R _{ S }= I =\frac{7}{35}=\frac{1}{5} A\)

Current through \(90 \Omega= I _{2}=\frac{15}{90}=\frac{1}{6} A\)

Current through zener \(=\frac{1}{5}-\frac{1}{6}=\frac{1}{30} A\)

Power through zener diode

\(P = VI\)

\(P =15 \times \frac{1}{30}=0.5 watt\)

\(P =5 \times 10^{-1} watt\)

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