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I had a question about a circuit.

First, consider this circuit, which is a parallel circuit and therefore easy to find the resistance of:

enter image description here

What happens when you add an additional resistor next to the battery like so:

enter image description here

Which part of the circuit is in series and which part is parallel? Is it possible to ascertain the resistance of this circuit at a glance? Or must one use Kirchoff's Laws, set up a system of equations to find the equivalent resistance?

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closed as off-topic by ACuriousMind Mar 27 '17 at 0:07

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  • $\begingroup$ R1 andR2 are in parallel and that parallel arrangement is in series with R3 and the cell. $\endgroup$ – Farcher Mar 26 '17 at 19:23
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The easiest way to solve this problem ist to use Thevenin's theorem for the open terminals which arise when you cut open the circuit where the resistance $R_3$ is inserted. It is easily seen that the Thevenin voltage will be the battery voltage $V_B$ $$V_{th}= V_B$$ and the Thevenin resistance is the parallel circuit resistance of $R_1$ and $R_2$ $$R_{th}^{-1}=R_{1}^{-1}+ R_{2}^{-1}$$ Thus the resistance of the circuit as seen from the battery is $$R=R_{th}+R_3$$ and the current flowing through the battery is $$I_B=\frac{V_B}{R}$$

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The answer above about redrawing the circuit really hits the nail on the head. The problem here (IMHO) is that when they ask these questions for homework or an exam they deliberately draw it in a way which is "silly" i.e. not how a sensible electronics engineer would draw it. To get round this practice drawing loads of circuits in the different ways possible which are all the same electrically. This gets your mind to recognize on a "silly" diagram what it is really supposed to be in the "sensible" diagram. After a while you will do this quickly in your brain without thinking about it.

Note this method won't work for much more complicated diagrams unless you are one of the rare people who have a photographic memory and great 3-D visual image capability in your brain.

The difficulty on these simple resistor diagrams is very common in students, I had it myself, and as well as re-drawing the diagrams I used to mess about with bagfuls of assorted resistors connecting them various which ways and measuring the result on a multimeter while trying to solve the physics homework problems. It's not really cheating unless you try to smuggle a bagful of resistors and a soldering iron into the exam room (NOT recommended !).

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The best thing you can do inorder to solve these kind of question is; assign potential across resistor logically.

  1. Resistor having same potential difference across them are in parallel.
    1. Otherwise, in series. Caution: Always first solve series and replace the whole series resistor with one resistor of effective resistance you measured and then go to calculate for parallel one.

Assume emf of battery to be anything like say, 'V'. So say positive terminal have potential V and negative have 0. So for the first case, as both the resistor have same P.D. V across them, hence they are in parallel. For second case, same thing we do but for R3 there will be a potential drop across it ( as we know there is always a potential drop across a resistor, if current flows through it) let say its 'x'. Therefore 1. PD across R1, 'x' 2. PD across R2, 'x' 3. PD across R3, 'V-x' So, R1 and R2 are in parallel, where as R3 in series. So add R1 and R2 as parallel and then add the resulting effective resistance in series with R3. enter image description here

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  • $\begingroup$ Your picture is not very readable. Can you take a better one? $\endgroup$ – ZeroTheHero Mar 27 '17 at 0:04

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