I'm new to this forum, and I hope to both contribute and seek knowledge from this forum. You may be disappointed, given my lack of Physics knowledge, but I will try my best.

Anyway, my question involves circuits. I understand how to resolve circuits in parallel and series, but I just don't get these. The original question was:

Choices (A), (B), (C), and (D) are attached below. Choice (E) is, as you might imagine:

Each of the circuits can be reduced to a single equivalent resistance by only using the rules for series and parallel resistors.

Would any kind soul be willing to explain to me how to combine these resistors and how to see the parallel and series aspects quickly and efficiently? Thanks!

To see how to combine the resistors, just attack each circuit one piece at a time: when you see resistors in series you know you can replace them by a single resistor; similarly, when you see resistors in parallel, they can also be replaced.

Start by redrawing each circuit, pushing the battery off to one side. As long as you don't change what connects to what, you can redraw the circuit in whatever way makes it easier for you to analyze.

If you need additional help, label the resistors in each diagram and we can walk you through a few examples.

To get through such problems label the nodes (points where 2 or more circuit elements meet) and/or the resistors. Then redraw the circuit to make it look less complicated to YOU. Once you think you're done, try to see which two (or more) resistors have the same potential difference (read below) across them, they are in a parallel combination...reduce them to an equivalent resistance right away. Draw the simplified circuit in stages to see whats happening.

All this may sound tedious but once you have some practice, such questions can be very painless. Later on if you pursue the subject of electric circuits, you will come across two techniques of simplification which usually make mincemeat of resistive circuits: Thevenin Theorem and Wye-Delta Transformations. But to get a point when you hardly need to worry much, it is necessary that you take the hard way first.

IDENTIFYING SERIES AND PARALLEL RESISTANCES:

If [tex]R_{1}[/tex] and [tex]R_{2}[/tex] are two resistances connected in series, ONLY one terminal of [tex]R_{1}[/tex] and one terminal of [tex]R_{2}[/tex] are connected together. You can visualize a series combination as a string which contains smaller masses stacked together. Each mass is placed in succession along a length but no two masses are linked together in a closed loop.

If the resistors are in series, the current through each resistor in the combination is equal (as can be shown using the Kirchoff laws). If two resistors are connected in parallel, they are connected in a closed loop. Hence, you can form a parallel combination from a series combination by joining together the 2 formerly unconnected free terminals of the resistors.

Please note that this is not a tactic to help you solve problems...you can develop your own methods with circuits and get familar with combinations only if you do problems and think along.

Doc Al and Vivek, thanks a bunch! I never realized that a closed loop meant that the two resistors were parallel. I guess that's the problem with my teacher not teaching us the fundamentals...

I'm still having trouble figuring out how to redraw a circuit diagram that's simpler. Would you mind the extra trouble to walk me through the process? I would really appreciate it. :)

Naturally, you picked the one example that does not allow simple reduction to a single equivalent resistance.

The way I would do these kinds of problems is simple minded: Look for unambiguous cases of resistors in series or parallel and then replace them with their equivalents. Keep doing that until you can't do any more replacing.

In circuit C: resistors 1 and 2 are clearly in series so they can be replaced by an equivalent resistor. That's it! All the other resistors do not form simple series or parallel combinations.

If you label circuit D, we'll have more to talk about.

Edit: My mistake: Circuit C can be reduced just like the others. See later post for solution. I plead temporary insanity.

Are you sure about that? The test that I got that problem from said that the answer was E: that all of the circuits could be reduced. :)

I actually think that I understand the other three now. I would approach one of these problems the same way that you do (simplifying existing parts), except that resistors in parallel don't appear that obvious to me. But with Vivek's tip of looking for closed loops, I think I can manage A, B, and D.

Oh, I didn't see that 3 & 5 are in parallel. Yep, I understand what you're saying. Thanks for checking that one again.

Just one last question...how do you identify two resistors that are in parallel, besides checking for a closed loop? While closed loops do allow me to identify resistors in parallel, it doesn't help much in simplifying it. For example, in problem C, I can see that 3 and 5 are in parallel, but it takes me a while to figure out where that equivalent resistor goes and how it fits into the circuit.

I'm not seeing your point here. Once you identify that two resistors are in parallel you can imediately redraw the circuit by replacing those two resistors by their equivalent. If between two points you see two parallel resistors (a closed loop), put the equivalent resistor between those same two points.

Could this be the issue: In circuit C, on the left of resistors 3 and 5 there is an obvious point where they connect. But on the right, it's not so obvious. There is a stretch of conductor that can be contracted to a point. (You can always redraw a circuit, stretching or contracting conducting paths as you see fit.) Once you "contract" it, then it's obvious that 3 and 5 are in parallel and between what points.

Just keep redrawing them before you write the equations...check each step (a solved problems book and/or a teacher can be a big boon as long as they don't spoonfeed us) yourself until you're convinced. Practice makes circuits perfect ;-)