Question 3.12: Find IB, IC, and vo in the transistor circuit of Fig. 3.41. ...

For the input loop, KVL gives

-4+I_{B}\left(20 \times 10^{3}\right)+V_{B E}=0

Since V_{B E}=0.7  V in the active mode,

I_{B}=\frac{4-0.7}{20 \times 10^{3}}=165  \mu A

But

I_{C}=\beta I_{B}=50 \times 165  \mu A =8.25  mA

For the output loop, KVL gives

-v_{o}-100 I_{C}+6=0

or

v_{o}=6-100 I_{C}=6-0.825=5.175  V

Note that v_{o}=V_{C E} in this case.