Calculating Node Voltages with KCl"

[TheRedDevil18]

Homework Statement

Homework Equations

The Attempt at a Solution

I'm not getting the correct answers, here are my equations

Node A:
80 = ix + i2
80 = Va/0.143 + (Va-(Vb+10))/0.2......1

Node B:
20 = iy-i2
20 = Vb/0.125 - (Va-(Vb+10))/0.2......2

Are these equations correct ?

 

[donpacino]

equation 1... i see a sign error
equation 2...same problem

 

[TheRedDevil18]

equation 1... i see a sign error
equation 2...same problem

I don't see where the sign error is

 

[donpacino]

There is an error with your I2 term. Look at the relationship between VB and the voltage source

 

[TheRedDevil18]

There is an error with your I2 term. Look at the relationship between VB and the voltage source

I don't know, Is it -10V ?, I'm confused with the signs

 

[NascentOxygen]

80 = Va/0.143 + (Va-(Vb+10))/0.2......1

You drop down from Vb by 10v to get to the 0.2 ohm resistor. Should be ...(Vb-10)...

If you draw an arrow from (-) to (+) on the battery, which you should do and label it +10V, you can see the drop in potential in going from Vb towards the 0.2 ohm. The bottom of the battery is 10v less than the top.

 

[TheRedDevil18]

You drop down from Vb by 10v to get to the 0.2 ohm resistor. Should be ...(Vb-10)...

If you draw an arrow from (-) to (+) on the battery, which you should do and label it +10V, you can see the drop in potential in going from Vb towards the 0.2 ohm. The bottom of the battery is 10v less than the top.

I still don't get the correct answers. I get Va = 5.65V and Vb = 7.56V
These are my equations
80 = Va/0.143 + (Va-(Vb-10))/0.2

20 = Vb/0.125 - (Va-(Vb+10))/0.2

 

[donpacino]

I still don't get the correct answers. I get Va = 5.65V and Vb = 7.56V
These are my equations
80 = Va/0.143 + (Va-(Vb-10))/0.2

20 = Vb/0.125 - (Va-(Vb+10))/0.2

that is what I got too. either we both made the same mistake, or the 'solution' is incorrect.

 

[NascentOxygen]

The answers accompanying the question in post #1 are consistent with the 20A source being directed downwards, so are not appropriate for the circuit as drawn.

 

[TheRedDevil18]

Ok, maybe the solutions are incorrect. I'm a bit confused with the second equation though, why is it Vb+10, because the 0.2 ohm resistor is still connected to the negative end of the battery. Shouldn't it be Vb-10 ?

 

[NascentOxygen]

It should be Vb - 10 for the reason I gave.

 

[TheRedDevil18]

It should be Vb - 10 for the reason I gave.

For both the equations ?

 

[NascentOxygen]

For both the equations ?

Yes, it defines the voltage at one end of that 0.2 Ω resistor.

 

[TheRedDevil18]

Ok thanks guys