A load resistor of `2 k Omega `is connected in the collector branch of an amplifier circuit using a transistor in common-emitter mode.The current gain`beta-50`.The input resistance of the transistor is `0.50 k Omega`. If the input current is changed by `50 mu A `,(a)by what amount soes the output voltage change by ,(b)by what amount does the input voltage change and (c) what is the power gain ?

Explanation

Given ` beta = 50, deltal_b = 50muA`
`V_0 = beta xxRG`
` = 50xx(2)/(0.5) = 200`
`(a) V_G = (V_0)/(V_1) = (V-0)/(deltal_(b) + R_(i))`
` = (200)/(50xx10^(-6)xx5xx10^(2)) = 8000V`.
`(b) delta V_1 = deltal_b xx R_i`
` = 50xx10^(-6) xx 5xx10^2`
` = 25xx10^(-3) = 25mV`
(c) Power gain ` = beta^(2) xx RG = beta^(2) xx (R_c)/(R_i)`
` = 2500xx(2)/(0.5)`
` = 2500xx(20)/(5) = 10^4.`