Circuit using maths and electricity laws/formulae

[SteliosVas]

Homework Statement

Okay so we are given this circuit. I do not want to be told step by step what the answer (or worked solution is) but I am wanting some guidance about how I can start/continue on from what is given to us.

We obviously need a understanding of Kirchoff's current law.

Our task is to construct the remaining equations and then solve for the currents i1 to i8 in terms of V and R.

Homework Equations

Okay so I know Kirchoff's current law that says that there is a conservation of charge. that is if there is only two wires one has current going in, the other equal going out. So the sum of all currents at node = 0A, and total charge in closed loop is zero. Also know Ohm's law is V=IR.

The Attempt at a Solution

[/B]
With that in mind we know:

0=i1+i2+i3+i4

and V=i6R+i7R+i8R

We can for instance combine the left triangle which has R and 2R and 2R all in parallel which should give us ((1/R) + (1/2R) + (1/2R) ) ^-1? Am i heading down the right path? And since that all combines to be one resistor that is parallel to the bottom left triangle which has a R and 2R again?

and so on and so forth? How do I do the current? Could it be like i2+i6=i7?

Thanks a lot!

 

[Qwertywerty]

Have you thought of using unkown voltages instead ? You would only need three variables then .

 

[SteliosVas]

Have you thought of using unkown voltages instead ? You would only need three variables then .

What do you mean? No I didn't think of it at all? Could you explain?

 

[Qwertywerty]

What do you mean? No I didn't think of it at all? Could you explain?

Could you by any chance label the junctions ? It is going to be a bit difficult to explain without it .

 

[SteliosVas]

Could you by any chance label the junctions ? It is going to be a bit difficult to explain without it .

You mean like this?

 

[Qwertywerty]

Yes , that's better . Now , let potential at A be zero , and V at D .

You will have three unknown voltages - V₁ , say , at E , V₂ at B , and V₃ at C .
Now write junction law at E , B and C , using voltages and resistances . You thus have three equations , three variables . Solve and find the variables' values .

Can you now write each individual current ?

Hope this helps .

 

[SammyS]

@SteliosVas ,

Use symmetry to simplify the circuit.

 

[SteliosVas]

What do you mean by symmetry? Sorry I do civil engineering not really too fond on electrical associated stuff.

So do you mean look at the left hand side of the circuit and than take the same for right? But doesn't it matter that the top right branch has a voltage source?

 

[SteliosVas]

Bump!

 

[willem2]

The circuit is symmetric about a line that goes through the middle of the voltage source, the middle of the circuit, and the middle of the resistance with i7 flowing through it. If you divide the voltage source in 2 voltage sources in series, both with potential difference V/2, and the resistor into two resistances R/2 in series, the 3 points on the symmetry axis all must have the same potential, If you connect them with wires, the rest of the circuit can be quickly solved with the rules for series and parallel resistors.

 

[SteliosVas]

Do you mean something like this?

 

[SammyS]

Do you mean something like this?

Yes, something like that.

You left out the 2R resistor between points E and C.

However, it's not two separate circuits the way you described it. If you split it that way neither part would be a circuit.Here's a different question to clarify what convention you may be using regarding current.

Which convention do you use for the direction of current flow?

A: Is it "conventional current" in which current is considered to be due to the flow of positive charge? It tends to flow in the circuit from the positive terminal of the voltage source to the negative terminal.

B: Is it "electron current" in which current is considered to be due to the flow of negatively charged electrons?​

 

[SteliosVas]

Yes, something like that.

You left out the 2R resistor between points E and C.

However, it's not two separate circuits the way you described it. If you split it that way neither part would be a circuit.Here's a different question to clarify what convention you may be using regarding current.

Which convention do you use for the direction of current flow?

A: Is it "conventional current" in which current is considered to be due to the flow of positive charge? It tends to flow in the circuit from the positive terminal of the voltage source to the negative terminal.

B: Is it "electron current" in which current is considered to be due to the flow of negatively charged electrons?​

Okay well I would consider conventional current, that is it flows out of the positive terminal and than back into the negative...

But I am still not sure, could I use matricies for this problem?

 

[SammyS]

Okay well I would consider conventional current, that is it flows out of the positive terminal and than back into the negative...

But I am still not sure, could I use matrices for this problem?

Yes, you could use matrices to solve this, but they are not needed.

The solution can be worked out with some simple parallel/series combinations.

Can you see that the current flowing from D to E must be equal to the current from E to A?

 

[SteliosVas]

Yes, you could use matrices to solve this, but they are not needed.

The solution can be worked out with some simple parallel/series combinations.

Can you see that the current flowing from D to E must be equal to the current from E to A?

Yes I understand that cause the value of the resistor is the same?

Okay so my understanding is this. Since the current flowing through D to E = E to A
Does that imply say that the current flowing from B to E is equal to E to C

Or did you say the flow because what ever is entering junction E from the right is exiting from the top?

Than B to C is parallel to C to D ?

I sort of understand that you can combine together. So you can treat that top right hand corner as a circuit in series with say the top left? But parallel to the bottom two?

Edit:
Okay so it asks to come up with the remaining equations for i₁ to i₈

Would I be wrong in saying the first two are?
2R_i1+2R_i2+2R_i3+2R_i4 = V

Making i₁+... i₄=V/2R

The second being:

R_i6+2R_i2-R_i7=V?

 

[SammyS]

Yes I understand that cause the value of the resistor is the same?

Okay so my understanding is this. Since the current flowing through D to E = E to A
Does that imply say that the current flowing from B to E is equal to E to C

One does imply the other, but they're both true because of the symmetry in the circuit.

Or did you say the flow because what ever is entering junction E from the right is exiting from the top?

This is not not enough reason to imply those equalities.

Than Then B to C is parallel to C to D ?

No. That is definitely not true.

The rest is not correct.

I sort of understand that you can combine together. So you can treat that top right hand corner as a circuit in series with say the top left? But parallel to the bottom two?

Edit:
Okay so it asks to come up with the remaining equations for i₁ to i₈

Would I be wrong in saying the first two are?
2R_i1+2R_i2+2R_i3+2R_i4 = V

Making i₁+... i₄=V/2R

The second being:

R_i6+2R_i2-R_i7=V?

I didn't read willem2's post #10 carefully enough, previously.

That may be a simpler way to analyze this circuit.

 

[SteliosVas]

Okay analysing it your way. How should I start? It's more the beginning that is the most difficult.

If you draw a line across the voltage source therefore anything lying on that line must (I think equal =0) so saying that one resistor will have value R/2 and the junction E a value and v/2?

Using that should I than analyse the top and bottom segments ?

 

[SammyS]

Okay analysing it your way. How should I start? It's more the beginning that is the most difficult.

If you draw a line across the voltage source therefore anything lying on that line must (I think equal =0) so saying that one resistor will have value R/2 and the junction E a value and v/2?

Using that should I than analyse the top and bottom segments ?

It is all the same voltage along that line. Maybe not zero.

At any rate, you can connect all along that line with a wire.

Which way do you want to analyze it?

The way willem2 suggests?

 

[SteliosVas]

I guess it doesn't matter which way I analyse it. My understanding of series and parallel circuits and related laws.

What I don't understand is what is meant by symmetrical and I guess what throws me is the voltage source

 

[SammyS]

I guess it doesn't matter which way I analyse it. My understanding of series and parallel circuits and related laws.

What I don't understand is what is meant by symmetrical and I guess what throws me is the voltage source

Does this help ?

 

[William White]

I guess it doesn't matter which way I analyse it. My understanding of series and parallel circuits and related laws.

What I don't understand is what is meant by symmetrical and I guess what throws me is the voltage source

Why don't you build the circuit using QUCs and see what is going on? You can probe parts of the circuit.
Then build the circuit Sammy suggests; then compare.

 

[SammyS]

Why don't you build the circuit using QUCs and see what is going on? You can probe parts of the circuit.
Then build the circuit Sammy suggests; then compare.

What the heck is a QUC?

I doubt that using a circuit simulator will help OP understand how to analyze such a circuit.

 

[William White]

http://lmgtfy.com/?q=QUCs

I think a circuit simulator might help. Because you can add bits as you go, check the currents, add a bit more, see what has changed. I think it's exactly like building simple circuits at school: you'd add lightbulbs and measure currents to get an intuitive grasp of what is happening. Sometimes being able to build something and measure what is going on is better than struggling with the maths, and struggling with an abstract concept of what goes where, which seems to be happening here.

 

[SammyS]

http://lmgtfy.com/?q=QUCs

I think a circuit simulator might help. Because you can add bits as you go, check the currents, add a bit more, see what has changed. I think it's exactly like building simple circuits at school: you'd add lightbulbs and measure currents to get an intuitive grasp of what is happening. Sometimes being able to build something and measure what is going on is better than struggling with the maths, and struggling with an abstract concept of what goes where, which seems to be happening here.

Good point.

 

[NascentOxygen]

I guess it doesn't matter which way I analyse it. My understanding of series and parallel circuits and related laws.

What I don't understand is what is meant by symmetrical and I guess what throws me is the voltage source

I would not have pursued "symmetry" quite as literally as the foregoing respondants.

Examining the pathways and resistor values, you can see that between point E and the top of the battery (i.e., the +V volts point) is exactly the same arrangement as between point E and the bottom of the battery (i.e., the 0 volts point). So, by symmetry, point E must be at a potential that is midway between +V and zero, meaning E must be V/2 volts.

Back to the circuit, you can see that between node B and the top of the battery is the same arrangement as is seen between node C and the bottom of the battery. Consequently, if we say node B is x volts less than the potential at the top of the battery, then, its corresponding node under symmetry is going to be x volts more than the potential at the bottom of the battery.

So, this leaves us with only one unknown in the whole circuit, viz., x. Once we know x we'll know the voltage at every node and hence the voltage across every resistor.

➤ Apply Kirchoff's and Ohm's Laws to determine x.

 

[SteliosVas]

http://lmgtfy.com/?q=QUCs

I think a circuit simulator might help. Because you can add bits as you go, check the currents, add a bit more, see what has changed. I think it's exactly like building simple circuits at school: you'd add lightbulbs and measure currents to get an intuitive grasp of what is happening. Sometimes being able to build something and measure what is going on is better than struggling with the maths, and struggling with an abstract concept of what goes where, which seems to be happening here.

That is an idea, but I would like to understand what is going on before I start using simulators lol!

I would not have pursued "symmetry" quite as literally as the foregoing.

Examining the pathways and resistor values, you can see that between point E and the top of the battery (i.e., the +V volts point) is exactly the same arrangement as between point E and the bottom of the battery (i.e., the 0 volts point). So, by symmetry, point E must be at a potential that is midway between +V and zero, meaning E must be V/2 volts.

Back to the circuit, you can see that between node B and the top of the battery is the same arrangement as is seen between node C and the bottom of the battery. Consequently, if we say node B is x volts less than the potential at the top of the battery, then, its corresponding node under symmetry is going to be x volts more than the potential at the bottom of the battery.

So, this leaves us with only one unknown in the whole circuit, viz., x. Once we know x we'll know the voltage at every node and hence the voltage across every resistor.

➤ Apply Kirchoff's and Ohm's Laws to determine x.

Okay so my understanding is you are saying basically the branch between B and the top of the batter is the same as C and the battery because the value of the resistor is R.
Since the voltage at E is 1/2 the branch at top will be the potential at E - a certain difference and bottom will be E + the certain difference.

 

[NascentOxygen]

Okay so my understanding is you are saying basically the branch between B and the top of the battery is the same as C and the battery because the value of the resistor is R.

No, I didn't attribute it to any single resistor. There is a complicated arrangement of resistors between B and both terminals of the battery, and we can't easily simplify it. But whatever it is, we can see there is an identical/symmetrical arrangement positioned between C and the battery terminals. And that's all we need to know---that from both points B and C the resistor-battery arrangement is identical [in symmetry]. This allows us to state---without further ado---that the +V battery terminal and point B have the same voltage difference as that between point C and the 0 volts end of the battery.

 

[SteliosVas]

No, I didn't attribute it to any single resistor. There is a complicated arrangement of resistors between B and both terminals of the battery, and we can't easily simplify it. But whatever it is, we can see there is an identical/symmetrical arrangement positioned between C and the battery terminals. And that's all we need to know---that from both points B and C the resistor-battery arrangement is identical [in symmetry]. This allows us to state---without further ado---that the +V battery terminal and point B have the same voltage difference as that between point C and the 0 volts end of the battery.

Thanks I eventually worked it out, I used a very hard way in the way of matrix but got it in the end

Thanks :)

 

[NascentOxygen]

Thanks I eventually worked it out, I used a very hard way in the way of matrix but got it in the end

But you will not always enjoy the luxury of being allowed to solve problems with symmetry using your "long way". These are favoured exam questions; they are solvable by all students, but only those who recognize the shortcuts of symmetry will reach the solution within the restricted time available in examinations.

I suggest that you attempt to solve it using the approach I outlined, for practice.