Hi I came cross this question on the internet, I work out the total I is 4.774amp on LTSpice then I can get the total resistance between the voltage is 2.0947ohm.

but I don't know how to work that out without LTSpice, can someone help me out?

 

I would rewrite the diagram to something easier to read by hand. You should start with looking resistor pairs. In this diagram, you have no values for any of your resistors so I am not sure how you came up with a value.

Here is the diagram I would work with. This should make it much easier to do by hand. Start with this, and if you still have problems I have this circuit solved out and can help you, but it would be better for you to attempt and learn from it first. Hint: R1 and R3 are in series and look up Thevenin Voltage to help you.

 

Thanks IronMod, I will look up Thevenin Voltage and try to solve it, and the values are R1=1, R2=2, R3=3, R4=4, R5=5

 

Here's what I would do. See the attachment.

Fig. 1 is the original circuit just drawn slightly different. R1, R3, and R5 form a delta configuration. I would convert it to a wye configuration as shown in Fig. 2.

R13 = R1*R3/(R1+R3+R5) = 3/9 Ω
R15 = R1*R5/(R1+R3+R5) = 5/9 Ω
R35 = R3*R5/(R1+R3+R5) = 15/9 Ω

Fig. 3 shows the series combination of R4 with R35 and R2 with R15.

Fig. 4 shows the parallel combination of the bottom pair of resistances.

My final answer is 155/74 = 2.09459 Ω.

 

I just realized that you can make a delta to wye conversion with R1, R3, and R5, or with R2, R4, and R5. You can also make a wye to delta conversion with R1, R2, and R5. All three options will make the circuit simple to solve. Once you master the technique of delta-wye and wye-delta conversions, it's really easy and a useful tool.

 

Thanks mlog, your method is nice and easy to understand.

 

yes, except that without formula sheet, nobody is going to remember equations for more than 5 minutes.

 

wonder what's the pros and cons on each convertion, anyone knows?

 

any method will give same results so simply use whatever works for you.
wye - delta formulas are very handy if you know them or have them accessible.
one can solve anything without them by using mesh or nodal analysis, just takes a bit more math.

 

In my opinion, nodal/mesh are more practical uses of analysis. They can be used for almost any circuit within reason, delta-to-wye only work for certain ones. I use voltage dividers and nodal/mesh the most.

 

Hi IronMod, can you post a solution with nodal/mesh method, I still don't know how solve it with nodal/mesh, since nodal/mesh is more practical, I think it's better for me to know it too

 

In my opinion, nodal/mesh are more practical uses of analysis. They can be used for almost any circuit within reason, delta-to-wye only work for certain ones. I use voltage dividers and nodal/mesh the most.

Well, delta-wye (or pi-tee, as it is sometimes called) will work here. On the other hand, you can write mesh equations, put them in vector-matrix form, and solve them using linear algebra.

You create an equation of the form V = Z * I. For this problem, you have three meshes with currents, i1, i2, and i3, which is a 3x1 vector; a 3x3 Z (impedance) matrix; and a 3x1 voltage vector with v1, v2, and v3. Then you invert the Z matrix and multiply it times the voltage vector in order to solve for the current vector. It works in this case, because one of of my currents is 4.77.

 

I do not mean to backtrack, and my apologies for this. But mlog is correct, for this particular circuit using a delta-wye transformation would have been the easiest way to do this. Sometimes in analysis, recognizing the quickest simplest way to do something is the best way, which is where I misled you and I apologize.

The solution is at my work, I will post it up tomorrow.

 

See this example, black loops are for KVLa, currents are blue and node voltages are red.
In this case we introduced bunch of unknowns but equations are simple and easy to write:
KCL@V1: I1+I2+I3=0
KCL@V2: I3=I4+I5
KCL@V3: I1+I5=I6
KVL@Loop1: V1-10V-V3=0
KVL@Loop2: 10V=I3*R1+I5*R2
KVL@Loop3: I2*R3=I3*R1+I4*R5
KVL@Loop4: I4*R5=I5*R2+I6*R4

we can also use
V4=I4*R5
V1=I2*R3
V3=I6*R4
to replace any voltage (like in KVL@Loop1):
I2*R3-10V-I6*R4=0

then we just need to solve system of simultaneous equations.
in this case those unknowns are just six currents so we don't really need all of the above equations.

one thing students are often asked to do is "supernode" analysis. this only cuts one step out of the process (not much of a saving). in this case DC source can be considered supernode but the same result is obtained from KCLsKCL@V1:

KCL@V1: I1+I2+I3=0
KCL@V3: I1+I5=I6

combining them we eliminate I1
I6-I5+I2+I3=0 (this would be obtained directly when using supernode).

The ultimate goal was to get equivalent resistance of the circuit.
That would be simply Ohms law (Req = 10V/I1). No need to memorize the wye-delta transformations.

sometimes problem is given as just resistor network (no source shown). the calculation is still the same, we just introduce our own test source Vtest (anything we like, such as Vtest=1V).

 

Thanks guys, I have learned two method to solve this kind of problem:

• delta wye conversion
• and, mesh/nodal analyse

this forum is great, I always get my question answer