Engineering Finding Voltage in a Transformer circuit

[eehelp150]

Homework Statement

Find Vo

Homework Equations

KVL, KCL

The Attempt at a Solution

Mesh equations:
-24 + I1(2-j)-I2(1.5j)=0
I2(6+4j)+10<30degrees-I1(1.5j)=0

I1=8.69541+3.59557j
I2=-1.00284+2.00908j

How would I find Vo?

 

[gneill]

How would I find Vo?

What's the potential difference across the output terminals comprised of?

 

[eehelp150]

What's the potential difference across the output terminals comprised of?

The 2 ohm resistor and 10<30V source?

 

[gneill]

The 2 ohm resistor and 10<30V source?

Yes. You know one of those (the voltage source). What's the potential difference across the resistor?

 

[eehelp150]

Yes. You know one of those (the voltage source). What's the potential difference across the resistor?

I2*2
-2.00568 + j4.01816

 

[gneill]

I2*2
-2.00568 + j4.01816

Okay. So you now know every potential change from the - terminal to the + terminal of Vo.

 

[eehelp150]

Okay. So you now know every potential change from the - terminal to the + terminal of Vo.

Would it be V(2ohm) + 10<30?

 

[gneill]

Would it be V(2ohm) + 10<30?

Yes, clearly it must be. You have learned about KVL and the summing of potential changes along a path between nodes, yes?

 

[eehelp150]

Yes, clearly it must be. You have learned about KVL and the summing of potential changes along a path between nodes, yes?

Are there any ways of verifying that the answer is correct?

 

[gneill]

Are there any ways of verifying that the answer is correct?

You could try to find a different solution method and confirm that you obtain the same results, although the loop current method that you've used is probably the most straight forward method and others might be more tricky to apply. Or you might use a circuit simulator to simulate the circuit. But there you'd need some experience dealing with transformers in a simulator.

For this problem I'd just go over the math and make sure it all looks good. Maybe do the problem again from scratch and make sure everything matches at the end. The critical part is obtaining the current in the second loop.

 

[eehelp150]

You could try to find a different solution method and confirm that you obtain the same results, although the loop current method that you've used is probably the most straight forward method and others might be more tricky to apply. Or you might use a circuit simulator to simulate the circuit. But there you'd need some experience dealing with transformers in a simulator.

For this problem I'd just go over the math and make sure it all looks good. Maybe do the problem again from scratch and make sure everything matches at the end. The critical part is obtaining the current in the second loop.

So I2*(-j2+4j) would NOT give the voltage of Vo?

 

[gneill]

So I2*(-j2+4j) would NOT give the voltage of Vo?

What is your reasoning for thinking it would?

 

[eehelp150]

What is your reasoning for thinking it would?

From the examples I did in nodal analysis.

If the voltage source was backwards (- on top, + on bottom), Vo would be V(2ohm)-10<30, right?

 

[gneill]

From the examples I did in nodal analysis.

You'd have to show the details of how you arrived at the particular value. I can't tell what you did by looking at a result only.

If the voltage source was backwards (- on top, + on bottom), Vo would be V(2ohm)-10<30, right?

Yes, in general if you change the polarity of a source it changes its sign in any KVL sum.

What you're doing in this instance is determining all the potential drops across the components along a path between two points where you wish to know their potential difference. The two points in question are where Vo is shown to be on your circuit. A handy path for that passes through the 10 V source and a 2 Ohm resistor.

So first you find the potential drop across the 2 Ohm resistor (the 10 V source is trivial). To do that you use the current flowing through the resistor. Pencil in the potential change polarity for that resistor on your circuit using your chosen direction for that current:

$i_2$ flows through that resistor from top to bottom, hence the +...- indication as shown. Then do your "KVL walk" along the path (indicated by the red arrow from a to b), summing the potential changes as they are encountered.

 

[eehelp150]

You'd have to show the details of how you arrived at the particular value. I can't tell what you did by looking at a result only.

Yes, in general if you change the polarity of a source it changes its sign in any KVL sum.

What you're doing in this instance is determining all the potential drops across the components along a path between two points where you wish to know their potential difference. The two points in question are where Vo is shown to be on your circuit. A handy path for that passes through the 10 V source and a 2 Ohm resistor.

So first you find the potential drop across the 2 Ohm resistor (the 10 V source is trivial). To do that you use the current flowing through the resistor. Pencil in the potential change polarity for that resistor on your circuit using your chosen direction for that current:
View attachment 108873

$i_2$ flows through that resistor from top to bottom, hence the +...- indication as shown. Then do your "KVL walk" along the path (indicated by the red arrow from a to b), summing the potential changes as they are encountered.

I assumed Vo was the voltage at node B (your picture) which is: V(2ohm) + 10<30 OR I2*(-2j+4j)

 

[gneill]

I assumed Vo was the voltage at node B (your picture) which is: V(2ohm) + 10<30 OR I2*(-2j+4j)

Yes, it's the potential at node b with respect to node a. I don't understand where I2*(-2j+4j) comes from; I don't see a path from a to b where that would be true.

 

[eehelp150]

Yes, it's the potential at node b with respect to node a. I don't understand where I2*(-2j+4j) comes from; I don't see a path from a to b where that would be true.

This is what I thought:

 

[gneill]

I see. But that is not a complete path from a to b. A complete path that includes those components would also include the transformer (both the secondary coil and the mutual inductance) and the 4 Ohm resistor. It has to be a complete, unbroken path.