In a common-emitter configuration, a transistor has `beta = 50` and input resistance 1 k `Omega`. If the peak value of AC input is 0.01 V, then the peak value of collector current is

Options

A. A)`0.01 mu A`
B. B)`500 mu A`
C. C)`100 mu A`
D. D)`0.5 mu A`

Answer

B

Explanation

Given, `beta = 50, V =0.01 and R = 1 k Omega = 1000 Omega`
Here, `beta = (i_c)/(i_b)`
`(i_c)/(i_b) = 50`
`(i_b)_("max") = (0.01)/(1000) = 1xx 10^(-5) A`
`(i_c )_("max") = (i_b)_("max") xx beta`
`(i_c)_("max") = (1xx 10^(-5) ) xx 50`
`(i_c)_("max") = 500 muA`