Using KVL as
-1 + (30K Ohm)(IB)+0.7=0
Base Current is found to be IB= 800uA,
I am confused how to find Vo and Io , please describe mathematically.

 

Hi,

Problem with your "30K", it is a 300k resistor.

Start by calculating the base current as you did but using the right resistor value, unless of course you intended to use a different value for that resistor.
Multiply by the Beta to get the collector current Ic, which is directed downward.
Use Thevenin and/or Norton to convert the battery and 20k to a current source in parallel with the 20k, we'll call it Ix.
The current through the two now parallel resistors is Ip=Ix+Ic, and remember Ic will be negative or simply subtracted from Ip as: Ip=Ix-Ic.
Multiply Ip times the parallel resistance of the 20k and 200k to obtain Vo.
Check to see that the solution voltage is actually possible with a 10v battery.

 

-1 + (300KOhm)(IB)+0.7=0 ==> (300KOhm)(IB)=1-0.7=300mV

 

You have two loops there.
You only did one loop. The easy one.
Now do the other loop.

 

View attachment 101214
Using KVL as
-1 + (30K Ohm)(IB)+0.7=0
Base Current is found to be IB= 800uA,
I am confused how to find Vo and Io , please describe mathematically.

Show your work for how you came up with Ib = 800 uA.

That's not correct, but we can't begin to tell you what you did wrong unless you show us what you did, period.

 

Hi,

Nice diagram by Bordodynov in post #3. That illustrates the electrotopological transformation to simplify the problem nicely i think. Unfortunately it looks like the OP may not be returning or is taking a long time to reply.