How can i find Vo voltage and Io current? This question is very complicated for me. Please help me.

 

The op amp will likely go to full voltage on either power supply rail, so it depends on the component being used. Very few op amps anywhere near the power supply rails.

You don't have a component for the op amp, you don't have power supply definitions.

In short, as stated the question is unanswerable.

 

Being a homework question, it is either ideal or a 341 .

 

Show us how you think the solution to the problem should be set up.

hgmjr

 

Find V- (inverting input voltage with respect to ground) and V+ (non inverting input voltage with respect to ground).

Then equate V- and V+ (ideal op amp) and you should find Vo with respect to Vin.

 

Using the hint Mik3, see if you can take a stab at solving the problem and post it here.

hgmjr

 

V+=(Vo*12k)/(12k+8k)=0.6*Vo , voltage divider rule

current through 4k and 16k is: assume current flows from Vo to 6V

I=(Vo-6)/(4+16)k=(Vo-6)/20k

thus

V-=6+I*4k=6+4k(Vo-6)/20k=6+(Vo-6)*0.2=6+0.2*Vo-1.2=0.2*Vo+4.8

assuming V-=V+

0.6*Vo=0.2*Vo+4.8

thus

0.4*Vo=4.8

thus

Vo=12V

Io=12V/10k=1.2mA

 

Millman's Theorem Solution.......

\(\small V_-=V_+\)

\(\frac{\frac{6}{4K}+{\frac{V_o}{16K}}}{\frac{1}{4K}+{\frac{1}{16K}}}=\frac{\frac{V_o}{8K}}{\frac{1}{8K}+{\frac{1}{12K}}}\)

\(\frac{6(16K)+4K(V_o)}{16K+4K}=\frac{12K(V_o)}{12K+8K}\)

\(\frac{6(16K)+4K(V_o)}{20K}=\frac{12K(V_o)}{20K}\)

\(\normalsize {6(16K)+4K(V_o)}={12K(V_o)}\)

\(\normalsize {8K(V_o)}={6(16K)}\)

\(\large {V_o}={\frac{6(16K)}{8K}}\)

\(\large {V_o}={6(2)}\)

\(\large {V_o}=12V\)

****************************************

\(V_+=\frac{12K(12)}{12K+8K}\)

\(V_+=\frac{12K(12)}{20K}\)

\(V_+=7.2V\)

****************************************

\(I_o=\frac{V_o-V_-}{16K}+\frac{V_o-V_+}{8K}+\frac{V_o}{10K}\)

\(I_o=\frac{12-7.2}{16K}+\frac{12-7.2}{8K}+\frac{12}{10K}\)

\(I_o=\frac{12-7.2}{16K}+\frac{12-7.2}{8K}+\frac{12}{10K}\)

\(I_o=0.0003+0.0006+0.0012=0.0021A\)

hgmjr

 

Hgmjr,

You are right about Io. I thought Io is the current through the 10k resistor only.

 

Thank you for your answers .