Calculate the output voltage of the circuit

[damien88]

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I am trying to calculate the output voltage of the circuit as viewed between the two nodes + - and then the equivalent resistance of the circuit looking into the two nodes.
I have suppressed the voltage source to obtain Req=30k//30k=15k ohms. Vout=-3v.
Suppressing the current source I obtain Vout=(30/30+30)x12v=6v.
Veq=-6v+3v=3v
Req= 15k ohms??

I don't have solutions to the questions and was hoping someone could tell me if this is correct or not before I attempt any more questions.

Thanks in advance

 

[berkeman]

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I am trying to calculate the output voltage of the circuit as viewed between the two nodes + - and then the equivalent resistance of the circuit looking into the two nodes.
I have suppressed the voltage source to obtain Req=30k//30k=15k ohms. Vout=-3v.
Suppressing the current source I obtain Vout=(30/30+30)x12v=6v.
Veq=-6v+3v=3v
Req= 15k ohms??

I don't have solutions to the questions and was hoping someone could tell me if this is correct or not before I attempt any more questions.

Thanks in advance

Do you really need to do anything with the sources? A voltage source is effectively a short circuit for impedance calculations, and a current source is an open circuit. That gives you the output impedance part of the answer. Then just write the KCL equation at the output node to calculate the output voltage...

 

[damien88]

That is what I have done by suppressing the voltage and current sources, replacing them with short and open circuits respectively. I am just not sure if I have re-drawn the circuit correctly to which I used to obtain an output of 3v.

 

[berkeman]

That is what I have done by suppressing the voltage and current sources, replacing them with short and open circuits respectively. I am just not sure if I have re-drawn the circuit correctly to which I used to obtain an output of 3v.

What do you mean re-draw the circuit? What equation did you use to get to Vout = 3V? (and where did the Vout = -3V come from?)

 

[damien88]

The question asks to suppress the voltage current sources individually to find the Vout. So after suppressing the voltage source I have both resistors in parallel i.e 15k ohms multiplied by the current so 3v. Made a mistake with direction of current source that's where I got -3v from.
I then suppressed the current source and took the Voltage across the 30k ohm resistor at the output, so (30k/30k+30k)x12volts. Summing the two output voltages is then 9volts.

Sorry if this is quite a basic example, electronics is not my strongest point.

 

[berkeman]

The question asks to suppress the voltage current sources individually to find the Vout. So after suppressing the voltage source I have both resistors in parallel i.e 15k ohms multiplied by the current so 3v. Made a mistake with direction of current source that's where I got -3v from.
I then suppressed the current source and took the Voltage across the 30k ohm resistor at the output, so (30k/30k+30k)x12volts. Summing the two output voltages is then 9volts.

Sorry if this is quite a basic example, electronics is not my strongest point.

Ah, I get it now. Yes, that is the correct answer. You can check that it is correct by just writing the KCL equation at the output node -- you only have one variable (Vout), so it's easy to solve the equation to check your work.

 

[damien88]

Thanks very much!