do not say pic isn't clear
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Hope this helps you mate
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In the circuit shown in the figure. the input voltage V1 is 20 V, Vbe - and Vce - 0 The values of 1g 1c and beta are given by
V i =V
BE+R
BiB ---------(1)
20=500×10
3×iB
iB =40μA
V o =V CE −RCiC ------------(2)
20=4×10
3iC
iC=5mA
β= IB
IC = 40μA
5mA =125
hope it helps u....!!
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