I want to find the total resistance between A and B. This is a HOTS mcq, which has the following options:

  • A: 45\$\Omega\$
  • B: 15\$\Omega\$
  • C: 5\$\Omega\$
  • D: 22.5\$\Omega\$

A book says that the answer is 5\$\Omega\$ as R1, R2 and R3 are parallel. But I cannot understand how can they be parallel? Our science teacher said that R1, R2 and R3 are in a series, so he said that total resistance R=45\$\Omega\$. The book that says that the answer is 5\$\Omega\$, does not have the method to find it out.

I tried that:

  1. All are in series :- 45\$\Omega\$

  2. 15\$\Omega\$ is parallel with 15+15+15\$\Omega\$ and with one more 15\$\Omega\$, which would turn into : 45/7\$\Omega\$, which is not possible.

  3. R3 is parallel with R1 and R2, as well as in a series with R2. R1 is parallel with R2 and R3, as well as in a series with R2 and R3. But then I cannot determine the method to get the answer.

So I am confused between 5\$\Omega\$ and 45\$\Omega\$. I think as the book says, the answer must be 5\$\Omega\$, but what is the method?

  • 1
    \$\begingroup\$ The first step in these types of problems is always to redraw the circuit in a way that isn't nonsensical. \$\endgroup\$
    – Matt Young
    Feb 20 '14 at 3:15
  • \$\begingroup\$ The book is correct. My professor explained it pretty well like this. Two things are in parallel when they both connect to the same nodes. Sounds simple, but that's all it is! Look at the left resistor, it connects node A with node B through the bottom short. The middle resistor connects node A and B through both shorts, and the right resistor connects A and B with the top short. \$\endgroup\$
    – krb686
    Feb 20 '14 at 3:20
  • \$\begingroup\$ Another way to think of it is current essentially always takes the path(s) of least resistance proportionally, right? So if you follow each path, you'll see there are 3 ways of traveling from A to B, each going through just 1 resistor. Current will NEVER travel through more than 1, since there is always an easier path. \$\endgroup\$
    – krb686
    Feb 20 '14 at 3:25

the book is pretty correct as the circuit is like this:


simulate this circuit – Schematic created using CircuitLab This circuit would be in series if there were no wires shorting the circuit and then net resistance would be 15 ohm.

  • \$\begingroup\$ While your explanation may be correct, your diagram doesn't match the original. His has 3 x 15R while yours has 3 x 5R. \$\endgroup\$
    – brhans
    Dec 29 '15 at 14:47
  • \$\begingroup\$ yeah you are right i just forgot resistances were of 15ohm however that will have no effect on concept that i have used in the circuit. thanks for correction brhans \$\endgroup\$
    – manav.tix
    Dec 29 '15 at 18:51


simulate this circuit – Schematic created using CircuitLab

Try redrawing the circuit with the 'A' connections in a bar at the top and the 'B' connections in a bar at the bottom (or use left/right).

They've deliberately made it confusing.

If you're redrawing, here's a set of guidelines that Olin wrote up:

Rules and guidelines for drawing good schematics


Yes, R1, R2 & R3 are parallel to each other & hence, resultant resistance is 5 ohms. While finding equivalent resistance between any two points, imagine that you're travelling from one point to the other. Let's consider here that we're travelling from A to B.

One end of R2 is short-circuited to A & other end is short-circuited to B. Same is the case for R1 & R3.

That's why they fall parallel to each other & total resistance is 5 ohms.


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