Find I 0. Assume ideal op-amp.

http://www.badongo.com/pic/4690906

I know that no current goes into an ideal op-amp. Does this mean that I 0 is zero by KCL?

 

I know that no current goes into an ideal op-amp.

No current goes into the inputs of an ideal op-amp. The output will source or sink as much current as needed for the ideal circuit.

 

OK, but how should I start here?

0.8 mA divides between the 10 and 6.4 ohms resistances. Some goes back to the current source through the 12 ohms resistance, while the rest goes back through the op-amp?

 

Since the op-amp i/p are high impedance, all of the 0.8 mA is through the 12K resistor. This tells the E drop across same resistor.

 

The output voltage is then 0.8 mA * 12 kOhms = 9.6 V.

The current through the 10 kOhm resistor is 9.6 V/10 kOhms = 0.96 mA, and the current through the 6.4 kOhms resistor is 9.6 V/6.4 kOhms = 1.5 mA.

The currents go from the 9.6 voltage to ground, so that I 0 is the sum of 0.8, 0.96 and 1.5 mA (3.26 mA), but in the opposite direction from that shown in the figure.

 

The output voltage is then 0.8 mA * 12 kOhms = 9.6 V.

There are 9.6V across the 12K resistor, yes. But is the - i/p of the op-amp at ground?