Find the supply voltage of a ladder circuit

[propvgvnda]

Yes the total resistance of the circuit is just R, (=11Ω) , so looking you have calculated right. I didn't ask you to digest information, just to redraw a diagram with the equivalent resistance R_CDH you found, well almost, by the calculation
R_CD + R_DHG = R/2 + R/2 = R .
If you had done that it would have been obvious you then had a pair of resistances in parallel whose equivalent resistance you can also calculate and then do another new diagram.Would then be evident you then have the same thing again. And it would now be evident that the voltage AF and the current at each node split in a very simple way and the answer to the problem would be immediately obvious.

I tried to redraw it. Got this

 

[propvgvnda]

You need to find the current in each successive upper (R/2) resister. The third one will be the requested current.
I don't know what current "at a node" means...currents flow through conductors and are confluent at nodes but are ill-defined there other than (what goes in) = (what comes out).

But I don't have the voltage to find it.

 

[haruspex]

But I don't have the voltage to find it.

You can find the V marked on your diagram (the voltage across CG) from the given current and resistances. From that, you can find the current in CG, etc.

 

[epenguin]

You know the total resistance of the circuit which turns out to be simply R (=11 Ω) .
If you knew the total current I_AB (=I_FE) required you would easily find the voltage V_AE.
You don't yet know the total current required you only know that required in branch HG. You would know that total if you knew the proportions in which the current splits at each junction.

That would be obvious if you did the diagrams completely.
If we call the original circuit Fig.1, then you have now done Figs. 3 and 5. Missing are Figs. 2 and 4 where you have summed the series pairs into a single equivalent resistance. On each diagram you should write the equivalent resistance of each element, far preferably in terms of R, and the letters for nodes.

If it seem to you an extravagrant labour to do four diagrams, I think if you did that it would become quite obvious to you how currents, and for that matter voltages, divide themselves at each junction, which would be better than making such slow progress.

 

[propvgvnda]

You know the total resistance of the circuit which turns out to be simply R (=11 Ω) .
If you knew the total current I_AB (=I_FE) required you would easily find the voltage V_AE.
You don't yet know the total current required you only know that required in branch HG. You would know that total if you knew the proportions in which the current splits at each junction.

That would be obvious if you did the diagrams completely.
If we call the original circuit Fig.1, then you have now done Figs. 3 and 5. Missing are Figs. 2 and 4 where you have summed the series pairs into a single equivalent resistance. On each diagram you should write the equivalent resistance of each element, far preferably in terms of R, and the letters for nodes.

If it seem to you an extravagrant labour to do four diagrams, I think if you did that it would become quite obvious to you how currents, and for that matter voltages, divide themselves at each junction, which would be better than making such slow progress.

I think the answer is 77 V

 

[epenguin]

OK in Fig 3 between the nodes which you should have labeled B, F you have the same voltage across, and you have two resistors of identical resistance (or, I prefer to say, conductance). What fraction of the total current will flow through one resistor and how much through the other?
(How much physics is required for that?)
Same question for CG in your Fig 1.

You know what current is required in HG so what does the above tell you about current required in AB?

Knowing the total current and the total resistance you can easily calculate total voltage required.

Because of the simple equalities of resistance, it is also easy to see at glance what the voltages are at each node.
There were simplifying relations of resistances in this case, but calculating the general case where all were different would not be much more difficult.

 

[propvgvnda]

OK in Fig 3 between the nodes which you should have labeled B, F you have the same voltage across, and you have two resistors of identical resistance (or, I prefer to say, conductance). What fraction of the total current will flow through one resistor and how much through the other?
(How much physics is required for that?)
Same question for CG in your Fig 1.

You know what current is required in HG so what does the above tell you about current required in AB?

Knowing the total current and the total resistance you can easily calculate total voltage required.

Because of the simple equalities of resistance, it is also easy to see at glance what the voltages are at each node.
There were simplifying relations of resistances in this case, but calculating the general case where all were different would not be much more difficult.

I think the current flow is the same through both resistors because they're equivalent.

 

[cnh1995]

I think the answer is 77

That would be correct if the input current (or supply current) was 7A.

You correctly found the equivalent resistance. That's part I.
Part II:
Now you need to use KCL and KVL to find the input current. This input current times the equivalent resistance will give you the input voltage U.
Refer the diagram in post #20 and start with the 7A in the bottom right branch. What is the voltage across resistance C?

Edit: My reply got posted late.

 

[epenguin]

I think the current flow is the same through both resistors because they're equivalent.

Yes.

 

[propvgvnda]

Refer the diagram in post #20 and start with the 7A in the bottom right branch. What is the voltage across resistance C?

77 V

Part II:
Now you need to use KCL and KVL to find the input current. This input current times the equivalent resistance will give you the input voltage U.

What node should I refer to write KCL and what loop do I choose for KVL?

 

[cnh1995]

77 V

Yes.
So what is the current through C?

Further, what is the current through the branch GF?

 

[propvgvnda]

Yes.
So what is the current through C?

Further, what is the current through the branch GF?

Do you mean through node C or C-resistor. Through C-resistor it's 7A. The current in GF must be also 7A.

 

[cnh1995]

Through C-resistor it's 7A.

Yes.

The current in GF must be also 7A.

No. How much current is entering the node G? The same amount of current will leave the node G.

 

[propvgvnda]

Yes.

No. How much current is entering the node G? The same amount of current will leave the node G.

I think HG current is 14 A. I(hg) + I(c) - I(FG) = 0

 

[haruspex]

I think HG current is 14 A.

You mean FG, yes? The current in HG is the given 7A.

 

[propvgvnda]

You mean FG, yes? The current in HG is the given 7A.

Yes, I meant FG. Well, is that correct?

 

[haruspex]

Yes, I meant FG. Well, is that correct?

Yes.

 

[cnh1995]

Yes, I meant FG. Well, is that correct?

Yes.
So what is the current through y? What is the potential at B?
Continue in this way towards left until you find the input current (current through EF).

 

[propvgvnda]

Yes.
So what is the current through y? What is the potential at B?
Continue in this way towards left until you find the input current (current through EF).

The current through y is 7A I think. How do I find potential?

 

[haruspex]

The current through y is 7A I think.

Then think again.
You have 7A in GH, 14A in FG. So what is the current in CG?
What is the current in DH and CD? So what is it in BC?

 

[propvgvnda]

Then think again.
You have 7A in GH, 14A in FG. So what is the current in CG?
What is the current in DH and CD? So what is it in BC?

CG is 7A. I don't know how to calculate the current is DH and CD.

 

[epenguin]

You know the current in HG and you don't know the current in DH??!

KCL

From going very slowly this seems to be going backwards now.

It is not helped by not having labelled nodes in your figures.

You should also write the currents as they are calculated on the figures.

I said the solution should be obvious once you drew the diagrams. In fact the whole solution and its argument from start to finish could be written in a single sentence! You have said how the current splits at C. Do I need to repeat that it splits in the same way at B? That not merely the principle but in this case even the resistance values are the same at B Fig 3 as at C Fig 1? So if you know the current at HG you know the current at AB. You don't need any of the potentials up to this point. You end up with Fig 4 in which you know resistances and the current AB, and you have to calculate the overall voltage. Circuit problems don't come much easier than Fig 4.