Question

# The transfer ratio of a transistor is $$50$$. The input resistance of the transistor when used in the common-emitter configuration is $$1\ K \Omega$$. The peak value for an alternating current input voltage of $$0.01$$ V peak is?

A
100μA
B
0.01 mA
C
0.25 mA
D
500μA

Solution

## The correct option is D $$500\mu A$$voltage gain $${ A }_{ v }=\beta \dfrac { { R }_{ c } }{ { R }_{ B } }$$peak value of colllector current$${ I }_{ c }=\dfrac { { A }_{ v }{ V }_{ m } }{ { R }_{ B } }$$$$\dfrac { \beta { V }_{ m } }{ { R }_{ B } }$$$$\dfrac { 50\times 0.01 }{ { 10 }^{ 3 } } =5\times { 10 }^{ -4 }A=500\mu A$$PhysicsNCERTStandard XII

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