# Multiple resistors not in series / parallel

1. Nov 22, 2012

### trollcast

1. The problem statement, all variables and given/known data
Circuit as shown: (sorry about the amazing paint skills)

The pd is being applied from A to B (C is in the centre of the circle)

(1) Determine the relations between the current by reversing V and using symmetry or by any other suitable method

(2) Calculate the total resistance accross AB

(3) Determine the magnitudes of the currents in terms of V

(4) Redraw the circuit with CD as the base.
If the pd is now applied across CD determine the resistance across CD

(5) Redraw the circuit with AC as the base. Explain why the resistance across AC cannot be calculated in terms of resistances in series / parallel

2. Relevant equations

3. The attempt at a solution

(1) Using Kirchoffs Current Law

$$I_1 + I_3 + I_5 = V/R_T \\ I_2 + I_4 + I_5 = V/R_T \\ I_3 + I_6 = I_4 \\ I_1 = I_2 + I_6 \\$$

Using Kirchoffs Voltage Law

$$-2I_1-2I_2+2I_5=0 \\ -2I_1-3I_6+I_3=0\\ -I_3-I_4+2I_5=0\\ 3I_6-2I_2+I4=0\\ -I_4+2I_5-I_3=0\\$$

(2) I can see that I can use the equations from (1) as simultaneous equations and therefore work out the resistance?

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2. Nov 22, 2012

### Simon Bridge

Did you try it and see?
7 unknowns and 9 equations certainly looks promising.

3. Nov 22, 2012

### Staff: Mentor

You can certainly attack the problem with KCL and KVL to arrive at a solution, but sometimes an appeal to symmetry can lead to a quicker and more elegant result.

Take a look at the 3R resistor. Suppose for a moment that it was removed from the circuit. How would the potential at D compare to the potential at C?

4. Nov 22, 2012

### trollcast

Without the 3R resistor the potential at C and D would be the same as the pd from A to C to B is split between a 1R and 1R resistor.

The pd from A to D to B is again split between a 2R and 2R resistor so its half the pd?

5. Nov 22, 2012

### Staff: Mentor

Right. What does that tell you about the current through the 3R resistor? Does the 3R resistor have any effect on the circuit?

6. Nov 22, 2012

### trollcast

It must equal zero ?

So not all the currents indicated in that circuit are actually flowing then?

7. Nov 22, 2012

### Staff: Mentor

Yup.
Zero is a perfectly good number

If a current through a branch happens to be zero you can remove that branch. Or, equivalently, if two nodes have the same potential you can join them with a short circuit (direct wire) without affecting the circuit operation. So here you can either remove the 3R resistor or replace it with a wire. That should make things easier to analyze.

8. Nov 22, 2012

### quantum2013

Could you please explain how symmetry would show a relation between the currents ?

for (3), would
i2=i1=$\frac{V}{4}$
i3=i4=$\frac{V}{2}$
be the correct answer to (3)?

Last edited: Nov 22, 2012
9. Nov 22, 2012

### Staff: Mentor

That's the idea, although technically you should include the R's in the expressions:

$i_1 = i_2 = \frac{V}{4R}$

$i_3 = i_4 = \frac{V}{2R}$

and don't forget to include expressions for the currents $i_5$ and $i_6$ .

10. Nov 22, 2012

### quantum2013

$i_5= \frac{V}{2R}$
$i_6=0$ because $i_6=-i_6$

Why are you including R's in the expressions if the question asks for magnitude of the currents in V?

11. Nov 22, 2012

### quantum2013

gneill how would you go about answering parts 4 and 5 of the question?

12. Nov 22, 2012

### Staff: Mentor

Because R is some resistor constant (a value that is specified only as a symbol in the problem statement). V is in volts, and Volts over Resistance yields current.

13. Nov 22, 2012

### Staff: Mentor

Why don't you make an attempt? We can't do your homework for you here...