Find the supply voltage of a ladder circuit

In summary: I believe it has the same current as the other branches.Yeah, I don't know. I don't understand how to identify what has which voltage and current.
  • #36
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.
 
Physics news on Phys.org
  • #37
epenguin said:
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.
 
  • #38
propvgvnda said:
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.
 
  • #39
propvgvnda said:
I think the current flow is the same through both resistors because they're equivalent.
Yes.
 
  • #40
cnh1995 said:
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
cnh1995 said:
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?
 
  • #41
propvgvnda said:
77 V
Yes.
So what is the current through C?

Further, what is the current through the branch GF?
 
  • #42
cnh1995 said:
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.
 
  • #43
propvgvnda said:
Through C-resistor it's 7A.
Yes.
propvgvnda said:
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.
 
  • #44
cnh1995 said:
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
 
  • #45
propvgvnda said:
I think HG current is 14 A.
You mean FG, yes? The current in HG is the given 7A.
 
  • #46
haruspex said:
You mean FG, yes? The current in HG is the given 7A.

Yes, I meant FG. Well, is that correct?
 
  • #47
propvgvnda said:
Yes, I meant FG. Well, is that correct?
Yes.
 
  • #48
propvgvnda said:
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).
 
  • #49
cnh1995 said:
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?
 
  • #50
propvgvnda said:
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?
 
  • #51
haruspex said:
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.
 
  • #52
You know the current in HG and you don't know the current in DH??! :oldsurprised:

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.
 
Last edited:
Back
Top