# Problem in Finding Answer

#### Ambitious_Probe

Joined Feb 7, 2016
2

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.

#### MrAl

Joined Jun 17, 2014
11,487
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.

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#### Bordodynov

Joined May 20, 2015
3,180
-1 + (300KOhm)(IB)+0.7=0 ==> (300KOhm)(IB)=1-0.7=300mV

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#### shteii01

Joined Feb 19, 2010
4,644
You have two loops there.
You only did one loop. The easy one.
Now do the other loop.

#### WBahn

Joined Mar 31, 2012
30,062
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.

#### MrAl

Joined Jun 17, 2014
11,487
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.