Electronics with Transistor (another problem)

[I like Serena]

I found this little electronics store... and they could fix my charger!
Me and my laptop are back in business again! Yeah! :!)

 

[Ouabache]

(from your post #5)
I'm just not sure about the last 2 clauses...is it all correct?

Hi FP !

In working through your exercise, I agree with your sol'ns for Ib, Ie, Vce, Vre. Good job !
However I don't believe you've correctly solved for Vb. Why do you feel Vb (the voltage at the base of the transistor) is only equal to IbRb??
Hint: if 12V is the potential (metaphorically, at the top of the hill) and you come part way down the hill to Vb. What should Vb equal?

I agree with ehild that answers are preferred in engineering, w/o so many zeros. (either scientific notation or using prefixes mA , \muA, mV, \muV). If you really like the zeros (for your peace of mind), you might state your answer e.g. 0.000043A = 0.043mA

There is something in this problem that will help you on the https://www.physicsforums.com/showthread.php?t=525613" (the one with the voltage divider R1 and R2, biasing network). In fact the labeling of the variables (eg. Vb, Rb) in the current problem is a hint at how useful it is to define them this way.

What I am referring to is your equation in the current problem that looks something like:
(a) : Vb - IbRb - Vbe - IeRe = 0
I recommend using this equation, in your earlier exercise. I promise, it will simplify your calculations immensely. (You may want to review how I defined Vb and Rb in https://www.physicsforums.com/showthread.php?t=525613&page=3,#35")

I miss your hand drawn-diagrams (photographed). Some of your new 'scans' are not bold or enlarged enough to read clearly. Your hand-drawn photographed images added a wonderful creative touch to your posts