Semiconductors

JEE Main level, numerical answer, medium

75 of 209

Question 75 of 209

The value of power dissipated across the zener diode (Vz=15V)\left({V}_{z}=15V\right) connected in the circuit as shown in the figure is x×10−1Wx\times {10}^{-1}W.

The value of xx to the nearest integer is _______.

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Answer: 5


Voltage across RS=22−15=7V{R}_{S}=22-15=7V
Current through RS=I=735=15A{R}_{S}=I=\dfrac{7}{35}=\dfrac{1}{5}A
Current through 90Ω=I2=1590=16A90\Omega ={I}_{2}=\dfrac{15}{90}=\dfrac{1}{6}A
Current through zener =15−16=130A=\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{1}{30}A
Power through zener diode
P=VIP=VI
P=15×130=0.5P=15\times \dfrac{1}{30}=0.5 watt
P=5×10−1P=5\times {10}^{-1} watt

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