Help with circuit problem

[bobhays]

EDIT: SOLVED Thanks for the help everyone.

 

 

 

 

Can I get some help with this problem?

 

 

Since the center node is connected to ground it has a voltage of 0.

That means (V1)/4 would be the current from V1 to the center node (and the same for V2 and V3).

 

My question is, the solution shows that the currents add to 0. Normally this would be true for a node since the current going in = current going out, but in this example the node is connected to ground. So wouldn't all the current going in go to ground? If it does go to ground then the equation isn't true since the currents wouldn't add to 0.

 

Can someone explain what I'm doing wrong?

 

EDIT: I understand what they did assuming that one of the voltages is negative, but idk how they assumed that one was negative.

[alexhelvetica]

If only I remembered anything I learnt in Physics this year!

[leadeater]

This may help you, been ages since I've done anything with kirchhoff's laws.

 

http://electronics.stackexchange.com/questions/108798/kirchhoffs-current-law-at-the-ground-node

[Stefan1024]

Sorry I don't have tome to solve it now. If you can wait 12 hours I will have a look.

But it's not that hard. You have 3 unknown voltages and 3 unknown currents. Figure out 6 non-depending equalations and solve them.

[bobhays]

Sorry I don't have tome to solve it now. If you can wait 12 hours I will have a look.

But it's not that hard. You have 3 unknown voltages and 3 unknown currents. Figure out 6 non-depending equalations and solve them.

Sorry my test is in 7 hours. I get that's how you solve it the only equation im having trouble with is

 

(V1)/4 + (V2)/10 + (V3)/5 = 0.

 

That implies that whatever current goes into the center node is leaving the center node (not through ground) ie. 0 current goes to ground.

If 0 current goes to ground that implies one of the nodes have a voltage lower than 0. I don't know how they came to that assumption.

[bobhays]

Sorry I don't have tome to solve it now. If you can wait 12 hours I will have a look.

But it's not that hard. You have 3 unknown voltages and 3 unknown currents. Figure out 6 non-depending equalations and solve them.

 

 

This may help you, been ages since I've done anything with kirchhoff's laws.

 

http://electronics.stackexchange.com/questions/108798/kirchhoffs-current-law-at-the-ground-node

 

Thanks for the link. So I see that sometimes ground is just there as a reference point and not actually a ground. If that were true than this equation makes sense.

 

How can I tell if the ground in the circuit is a symbolic ground or an actual ground? And would it make a difference? because the way I approach the problem is different if there is an actual ground or a symbolic one.

[Stefan1024]

Thanks for the link. So I see that sometimes ground is just there as a reference point and not actually a ground. If that were true than this equation makes sense.

How can I tell if the ground in the circuit is a symbolic ground or an actual ground? And would it make a difference? because the way I approach the problem is different if there is an actual ground or a symbolic one.

It is an actuall GND. But it has only one connection to your circuit. To allow a cuerent to flow you need a closed loop so you need at least two connections to GND.

[Ariito]

Have you've tried doing mesh analysis?

[leadeater]

In this case I think it is there to show 0 volts. And unless this is a trick question you cannot break kirchhoff's laws.

 

Try not to let what we say influence you too much, trust what you have learnt and in the math.

[DigitalHermit]

Redraw the circuit first... it'll provide a better view of the problem...

 

also that form is a common trap used by our instructors to fool us into wasting time with the wye-delta equations...

 

then solve it with mesh/node analysis...

[bobhays]

It is an actuall GND. But it has only one connection to your circuit. To allow a cuerent to flow you need a closed loop so you need at least two connections to GND.

So what's the point of it being there?

[Kalle Hänninen]

So what is the problem. You have 3 sources connected in one ground, which is shared by v1,v2,v3. Those measurements are measurements and idealized measurements do not affect the circuit, so you're left with a star.

[Stefan1024]

So what's the point of it being there?

It don't have to. But it's nice to give a reference so the circutit isn't floating.

[Kalle Hänninen]

Thanks for the link. So I see that sometimes ground is just there as a reference point and not actually a ground. If that were true than this equation makes sense.

 

How can I tell if the ground in the circuit is a symbolic ground or an actual ground? And would it make a difference? because the way I approach the problem is different if there is an actual ground or a symbolic one.

Ground is ground. Every ground is same ground. They may not be connected, but they could be and nothing would happen.

[Kalle Hänninen]

It don't have to. But it's nice to give a reference so the circutit isn't floating.

 

It need's to be there. You can't separate + and -, so those batteries are connected to ground.