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JEE MAIN 2020
09-01-2020 S2
Question
The circuit shown below is working as a 8 V dc regulated voltage source. When 12 V is used as input, the power dissipated (in mW) in each diode is; (considering both zener diodes are identical) __.
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Solution
$\begin{aligned} & i=\frac{(12-8)}{400}=10^{-2} \mathrm{~A} \\ & P=4 \times 10^{-2}=40 \mathrm{~mW}\end{aligned}$
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