Reducing Circuits to Their Simplest Form

[sweetdion]

Homework Statement

Which of the following circuits could be reduced to a single resistance and a battery by determining the equivalent resistance using combinations of resistors in series and parallel?

A. II and IV only
B. I and II only
C. II and III only
D. I and III only
E. I and IV only

Homework Equations

Parallel: 1/Req = 1/R1+1/R2+...
Series: Req= R1+R2+...

The Attempt at a Solution

I think 3 and 4 can be eliminated because it seems impossible that you can reduce those to just one resistor.

So I'm thinking the answer is I and II only, but I'm unsure.

 

[Melawrghk]

Take a closer look at 4. Redraw it (hint: it'll have 3 parallel resistor branches).
For 2, are you planning to use the delta to Y transformation?

 

[sweetdion]

Take a closer look at 4. Redraw it (hint: it'll have 3 parallel resistor branches).
For 2, are you planning to use the delta to Y transformation?

I don't even know what you mean by the delta to Y transformation

 

[Melawrghk]

I don't even know what you mean by the delta to Y transformation

Don't worry about it then :) It's a transformation that would allow you to change circuit 2 into battery+resistor circuit. Since you said you thought it could be solved, how did you plan on reducing on?

 

[sweetdion]

well the resistors on the outer edges of the diamond are in series. You can reduce those to one resistor each. Then you have 3 resistors in parallel which you can reduce...i think

 

[Melawrghk]

well the resistors on the outer edges of the diamond are in series. You can reduce those to one resistor each. Then you have 3 resistors in parallel which you can reduce...i think

For circuit 2? That wouldn't be correct. Every resistor is connected to nodes (point where 3 or more wires meet) at both of its ends. You can only claim that resistors are in series if they do not have a node inbetween them.
---R1----R2---- Resistors 1 and 2 are in series

---R1---R2---- Resistors 1 and 2 are NOT in series
''''''''''''''|
'''''''''''''R3
'''''''''''''|

That is however the correct approach for circuit 4.

 

[sweetdion]

Okay, So I see what your saying. The correct answer would be E. I and IV, unless I've got something wrong out of your explanations...