Hello,
Can anybody show me how to solve this problem (Value of R2)?

 

Best solution for me is using Kirchhoff´s law:
1) Sum of currents in one joint equals 0
2) Sum of voltages in one loop (don´t know the right term) is equal to zero.

Look at www.allaboutcircuits.com for explanation

 

Check Volume 1 Chapter 6: Divider Circuits and Kirchoff's Laws.

Dave

 

Once you have reviewed the material contained in the link that Dave has posted, then give the calculation a whirl and then post your calculations and your results. We will be happy to provide you a bit of guidance if needed.

hgmjr

 

I came up with 13K for R2.
Vo = 0.25ma * 5K +0.7V
Vo = 1.95V
I1= I2 + ID
I1 = 12V/30K =0.4mA
ID = 0.25mA (Given)
I2 = 0.4-0.25 =0.15mA
V0 = 1.95V
R2 = 1.95V/0.15mA = 13K

 

The part about 12v/30k doesn't look right; the voltage drop across the 30k can't be all of the 12v, it's got to be something less than that, seems like about 9 or 10 volts, because you've got to subtract the voltage drops caused by all of the 3 remaining components. I got 40678, but I don't have time to check it before I leave---Sorry, Rich

 

Vo = 12-1.95 = 10.05V

I1 = 10.05/30K = .335mA

I2 = 0.335-.25 = .85mA

R2 = 1.95/0.85 = 22.94K

 

Vo = 12-1.95 = 10.05V

I1 = 10.05/30K = .335mA

I2 = 0.335-.25 = .85mA

R2 = 1.95/0.85 = 22.94K

Right..... except 0.335-0.25=0.085 and 1.95/0.85=2.2941 so 1.95/0.085=22.941

 

Not a bad effort, annu007.

I used Millman's Theorem and arrived as 22.941K....

hgmjr