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schematic

simulate this circuit – Schematic created using CircuitLab

Original schematic

Could anyone offer some insight on how to solve this circuit? What I've been doing so far, is only producing a wrong answer... I've been getting that V_A = 15 (which is wrong). Also I think V+ is = 0 since the node below 3 ohm is grounded... Any help on how to start solving this?

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    \$\begingroup\$ Please show the steps of your calculation so someone will be able to teach/point out where you are stuck. \$\endgroup\$ – shimofuri Apr 2 '13 at 17:03
  • \$\begingroup\$ Ok! So, what I did at first was: V+ = 0, the current 5Am is going through 3 ohm resistor => Va- 0 = 3 *5 = 15. Afterwards ((V_-) - 15)/ 2ohm + (V_0 -V-)/3ohm = 0 \$\endgroup\$ – user65165 Apr 2 '13 at 17:06
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    \$\begingroup\$ Add junction dots and component designators! \$\endgroup\$ – Olin Lathrop Apr 2 '13 at 17:27
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    \$\begingroup\$ So then fix it. What matter here is how you present it here. \$\endgroup\$ – Olin Lathrop Apr 2 '13 at 17:41
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    \$\begingroup\$ Off-topic retoric question: I understand this is just for learning electronic circuit analysis and one of the goals is of course to be able to work with nice round numbers, but be aware that in practice the average opamp can sink/source nowhere near the 2A that it does here. I mean, what is wrong to get students used to work with practical values? OK, I said my thing, now solve the circuit :o) \$\endgroup\$ – jippie Apr 2 '13 at 18:22
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Here is some insight. The + input of the op-amp is grounded, and the negative feedback action has the effect of keeping the - input of the op-amp at the same voltage as the + input. So the - input is a virtual ground.

You can solve this by pretending that the op-amp is not there, and then solving the remaining network of resistors and current source by finding the value of \$V_o\$ which is necessary for the junction between the \$3\Omega\$ and \$2\Omega\$ resistors to be at zero volts.

The easiest way to do this is to pretend that the junction is actually grounded (not simply at 0V potential). Determine how much current is dumped into that ground through the \$2\Omega\$ resistor. But then, recognize that there is no ground there and so the current flowing toward that node actually returns via the \$3\Omega\$ feedback resistor.

The current across the feedback resistor gives you a straightforward way to determine the voltage across that resistor, which directly determines \$V_o\$.

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First, show junction dots in your schematic. It is ambiguous without junction dots whether the bottom of the current source and the + input are also connected to ground, or just the bottom end of the first 3 Ω resistor (Argh, use component designators too). I'll assume they are all connected to ground.

Look at the current source and resistor accross it. Can you convert that to something more directly usable? Hint: If you get stuck look up what Thevenin and Norton had to say.

Once you have done the substitution above, you have a basic inverting amplifier. That should be solvable just from inspection at that point. If you are still stuck there, show the updated circuit (with junction dots and component designators) and we can discuss more from there.

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Convert the current source to voltage source as shown below. Now it's a inverting amplifier.

schematic

simulate this circuit – Schematic created using CircuitLab

The solution is here:

Vo = -Vin * ( Rf/Rin )

Vo = -(-15) * (3/5) = +9 V

So the answer is +9V

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    \$\begingroup\$ Durgaprasad made correct answer, only slight change is needed. Since current flowwing from ground into R3, input voltage is -15V, and output is +9V Cheers, Patrick \$\endgroup\$ – Patrick Chung Apr 3 '13 at 8:21
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You can also aply a current divider.

First V- =V+. Then V- = 0 V. It's a virtual ground. And also through V+ and V- the current is nearly 0. So you the current that goes from the upper node of R_3 to V- is the same that goes from V- to Vo.

You can find the current just analysing this circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

$$I_{R1} = 5 * \frac{3}{3+2} $$ $$ I_{R1}=3A $$

Then, Vo will be:

$$V_o=R2*I_{R1}$$ $$V_o=3\Omega * 3 = \pmb{9V}$$

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