Semiconductors

JEE Main level, single correct, medium

37 of 209

Question 37 of 209

In the following circuit, the reading of the ammeter will be (Take Zener breakdown voltage =4 V)=4 \mathrm{~V})

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Answer: (C) 10 mA


V1=400100+400×12 V=45×12=485 V\mathrm{V}_1=\frac{400}{100+400} \times 12 \mathrm{~V}=\frac{4}{5} \times 12=\frac{48}{5} \mathrm{~V}
here, V1>Vz,(Vz=\mathrm{V}_1 \gt \mathrm{V}_{\mathrm{z}},\left(\mathrm{V}_{\mathrm{z}}=\right. Zener Voltage )) So, Zener breakdown will be take place So, voltage across 400Ω400 \Omega will be 4 V
I=4400 A=1100 A=10 mA\mathrm{I}=\frac{4}{400} \mathrm{~A}=\frac{1}{100 \mathrm{~A}}=10 \mathrm{~mA}

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