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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