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The transfer ratio \beta of a transistor is 50. The input resistance of the transistor when used in common – emitter configuration is 1k. The peak value of the collector AC current for an AC input voltage of 0.01 V peak is

Option: 1

\mathrm{100\mu A}


Option: 2

\mathrm{250 \mu \mathrm{A}}


Option: 3

\mathrm{500\mu\, A}


Option: 4

\mathrm{800 \mu \mathrm{A}}


Answers (1)

best_answer

Given that,

\mathrm{V_1=0.01 volt }

\mathrm{R_{\mathrm{l}}=1 \mathrm{k} \Omega=10^3 \Omega }

\mathrm{\therefore I_b=\frac{V_1}{R_1}=\frac{0.01}{1 \times 10^3}=0.01 \times 10^{-3} \Omega=0.01 \mathrm{~mA} }

\mathrm{Further, \mathrm{I}_{\mathrm{c}}=\beta \mathrm{I}_{\mathrm{b}}=50 \times 0.01 \mathrm{~mA}}

\mathrm{=0.5 \mathrm{~mA}=500 \mu \mathrm{A}}

Posted by

Suraj Bhandari

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