Question

I'm assuming my fantastic instructor wants us to solve for I1, Ic, and Ib. Either way, your guess is probably better than mine.

Updated: No supply voltage given by professor so most classmates just picked a voltage.

LEnMS .

Answer 1

We found this as -

We have let's common supply voltage used for BJT transistor implies 12 v by voltage divider, voltage across 10k implies 12 times 10/15 implies therefore I_1 implies 12 - 8/5 k implies 4/5k implies 0.8 mA I_2 implies 8/10 k implies 0.8 mA we can redraw circuit for Ib_1 so, v_th implies & R + h implies 10 k // 5k implies 3.33 k Ohm kvl to BE loop, therefore I_B implies 8 - 0.7/R_th + (c + B) R_E (because I_E implies (c + B) I_B) implies 7.3/(3.33 + 201 times 1) k therefore I_B implies 35.72 mu A & I_C implies beta I_B implies 35.72 times 10^- 6 times 200 therefore I_C implies 7.14 mA.