I am a second year mechanical engineering student. I am repeating an exam in Electronic Circuits. I have the exam on Thursday and am working through the old exam paper. I am having a few problems. I would really appreciate some help.

1) op am question-find v1 and v2

Attempt:
If the the voltage across the two inputs must equal 0. The V+ is 2v, the src beside it. Then must V- be also 2v. V1 already must equal 5v, so therefore for V- to be 2v then v2 must be 3v? If this correct? Looking back at this i don't think it is correct as we are talking about voltage, and it has to be across 2 points.
However if each input currents has to equal 0. Then i- must equal 0. The voltage V1 is equal to 5v, therefore i1 must equal .2amp, then i2 must equal .2 as well? Giving a V2 of 1?

2)Norton equivalent question

Attempt:
To get a Norton equivalent i must short the terminals and see what current goes through them. How do I calculate this current? Is it 7amps? What affect does the 1ohm resistor have?


3) Max Power Question

How do i find the Thevin equivalent resistance? Isn't it the resistance in the circuit "looking back" from the terminals and shorting the voltage src? Then is 177.77ohm correct?

4) current to voltage amp question

Attempt:
if i1 is 2mA. Then 5(i1)=.01, therefore using voltage division Vo is equal to .005 volts? I thought this was an amplifier?

 

How is V1 equal to 5V?
V1 is across the resistor, while the 5V is the source (different points).
You're right about V-, so the voltages on either side of your resistor are 5V and 2V.

I can't tell whether you're doing it incorrectly or getting the names confused.
V1 is across the resistor, which is the difference of the sources 5V and 2V.
V2 is across the second resistor, which has the same current as the first. Since the resistor is 5x larger the voltage must be 5x larger.

 

Thanks, no i was incorrect. I wrongly thought that 5v was across the first resistor. That question makes sense now.

 

Any chance someone cold help with the other questions? I am a bit stuck here.

 

I think with question 2, if you short the output terminals, there will be two amps going through the 1ohm resistor. This is because a current source provides a constant current regardless of voltage across it.

 

I think with question 2, if you short the output terminals, there will be two amps going through the 1ohm resistor. This is because a current source provides a constant current regardless of voltage across it.

Does the rest of the circuit not affect the Norton equivalent current at all then? Is it just 2 amps? Or do I also add/subtract the other current source?

 

For the Norton equivalent you need to find the short circuit current and the resistance.

Because there are no dependent sources you can find the resistance by just opening all the current sources and combining resistors.
You get:
(2 || 2) + 1 = 2 ohms.

The current however, is divided. You can combine the two current sources to be 3 A because parallel current sources add and these two are opposite polarities.
For the resistors 2 || 2 = 1. Now you're left with two 1 ohm resistors in parallel, with the short circuit current being in one of them. That gives you half, or 1.5 A.

 

Thanks a million Ghar, I was trying to do too many things at once. Should have slowed down and re-wrote the circuit bit by bit like you did. I understand it now, just needed to break it down.

 

Is this correct for finding out the Theivin equivalent Resistance in question 3? If i short the voltage source the rewrite the circuit as follows, i get 177.778 ohm. Is this correct? Also is this equal to the value of the load resistor applying the max power theorem? Or do I also have to work out the equivalent voltage?

 

Yep, that's how you get the resistance.

You should find the voltage though to make it easier to find the maximum power.
It's easier to find that Thevenin voltage than solving the circuit with a load.

You can easily find the voltage at either output node by the voltage divider rule. The Thevenin voltage is the difference between those.