What is the total resistance of the part of the circuit shown in the picture?

[lendav_rott]

Homework Statement

Attached diagram. The resistors don't have any concrete resistances.
Need to express the total resistance of the part of the circuit shown in the picture.

Homework Equations

Series - R= ƩR_n
Parallel - R = (ƩR_n^-1)^-1

The Attempt at a Solution

R1 and R2 make a series - so the resistance of it would be R1+R2 and same with R4+R5.
And using that approach I would have 3 parallel connections.
R1+R2
R3
R4+R5
And compute the total resistance with the parallel formula...but, something just doesn't feel right.

 

[SammyS]

Homework Statement

Attached diagram. The resistors don't have any concrete resistances.
Need to express the total resistance of the part of the circuit shown in the picture.

Homework Equations

Series - R= ƩR_n
Parallel - R = (ƩR_n^-1)^-1

The Attempt at a Solution

R1 and R2 make a series

No they don't. They do not have the same current flowing through them.

Neither are R₄ and R₅ in series.

- so the resistance of it would be R1+R2 and same with R4+R5.
And using that approach I would have 3 parallel connections.
R1+R2
R3
R4+R5
And compute the total resistance with the parallel formula...but, something just doesn't feel right.

This circuit, as it is, cannot be analyzed by means of series/parallel methods.

Either use Kirchhoff's Circuit Laws, or do something called a Y-Δ transformation on R₁, R₃, R₄ or on R₂, R₃, R₅. (After such a transformation you can do series/parallel analysis.)
http://en.wikipedia.org/wiki/Y-delta_transform

 

[lendav_rott]

Right, so I have a question.
Using the 1st diagram, If i let a current flow through the network, how do I determine the direction of the flow?
According to Kirchhoff the currents' summarum has to be 0. I have no way of showing that if I don't know in which direction the current flows in the 2 parts of the network, namely R1 R3 R4 and R2 R3 R5

Let's take the node that connects R1 R3 and R2. If a current I goes to the node, then it would have to split to 2 currents, say I3 and I2 and Kirchhoff says that I + I3 + I2 = 0. And again , what is the positive direction of flow, or do I assume one myself and do everything according to that assumption? Can I do it?

EDIT: Scratch all of what you just read. Thank you for mentioning the Kirchhoff's laws, I looked into it and found a very simple method called the Thevenin's Theorem to tackle such problems.

 

[CWatters]

If i let a current flow through the network, how do I determine the direction of the flow? According to Kirchhoff the currents' summarum has to be 0. I have no way of showing that if I don't know in which direction the current flows in the 2 parts of the network

It doesn't matter.

What you do is make an arbitrary decision about the direction of the current in each resistor and mark it on the diagram. This will determine the direction of the voltage drop through each resistor which you also mark on the diagram. These two have to be consistent with each other.

When you write the KCL equations they will typically be of the form..

(V1-V2)/Rn + (V3-V4)/Rm ... etc... = 0

where V1 and V2 are the voltages at each end of Rn.

The important thing is that the order (eg is it V1-V2 or V2-V1) is determined by the voltage drop on your diagram.

When you solve the equations some of the voltage drops and currents may turn out to be negative. That just means your arbitrary decision on the direction was wrong but it doesn't effect the results.

 

[CWatters]

Let's take the node that connects R1 R3 and R2. If a current I goes to the node, then it would have to split to 2 currents, say I3 and I2 and Kirchhoff says that I + I3 + I2 = 0

Actually no that's wrong. Your description implies you are defining I as flowing into the node and I2 and I3 as exiting the node. Let's define current into a node as +ve. In that case the equation is...

I + (-I2) + (-I3) = 0 ......(1)

or

I - I2 - I3 = 0

Confused? Imagine a node with just two wires. If the current into the node is I1 and the current leaving is I2 then clearly..

I2=I1
and so
I2 - I1 = 0

Suppose you define both currents as flowing into the node. Then

I2 = -I1
and so
I1 + I2 = 0