CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
0.01 mA
loader
C
0.25 mA
loader
D
500μA
loader

Solution

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

Physics
NCERT
Standard XII

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
Same exercise questions
View More


similar_icon
People also searched for
View More



footer-image