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The given circuit

All resistance are equal in value, \$R1 = R3 = R4 = R5 = R6 = 1~\Omega\$

My calculations are as follows:

  1. R1 and R3 are in parallel => \$R13 = 2~\Omega\$
  2. R5 and R6 are also in parallel => \$R56 = 2~\Omega\$
  3. R13, R4 and R56 are in a series => \$R_s = 5~\Omega\$
  4. \$R_{eq} = \dfrac{1}{R_s} = \dfrac{1}{5} = 0.2~\Omega\$

I am getting a total resistance equal to \$0.2~\Omega\$, but my teacher said that the correct answer is \$\dfrac{1}{4} = 0.25~\Omega\$. What am I doing wrong?

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  • \$\begingroup\$ 1 ohm in parallel with 1ohm is not 2 ohm \$\endgroup\$ – M.Ferru Apr 26 '17 at 15:39
  • \$\begingroup\$ What is the calculation for resistors in parallel? It often helps to re-draw these circuits to more easily see what the actual connections are. \$\endgroup\$ – Peter Bennett Apr 26 '17 at 15:43
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    \$\begingroup\$ Unless I made a mistake, both you and your teacher are wrong. (Your algebra is a little strange too.. eq 4. how can R= 1/Rs the units don't work!) \$\endgroup\$ – George Herold Apr 26 '17 at 16:08
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Well, you have the good method but, it seems you have some difficulties to find the equivalent resistor of 2 resistors in parallel.

$$ 1 / Req = 1/R1 + 1/R2 $$ $$ 1 / Req = (R2 + R1) / R2R1 $$ $$Req = R2R1/(R1 + R2)$$

If R1 = R2 = 1 ohm, Req is 0,5 ohm. I let you finish your exercice ;)

schematic

simulate this circuit – Schematic created using CircuitLab

Isn't it clearer?

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  • \$\begingroup\$ Yes, I made an error with formula but still, I think I don't make the right calculations for total resistance. R13 and R56 are indeed equal to 0.5 ohms each and when I add them with the last resistance, R4 the resulting total resistance is 2 ohms, (0.5 + 1 + 0.5). \$\endgroup\$ – Vrabii Daniel Apr 26 '17 at 15:50
  • \$\begingroup\$ Have you redraw every step of the calculation? I won't give you the answer, but just help you to find it ;) \$\endgroup\$ – M.Ferru Apr 26 '17 at 15:53
  • \$\begingroup\$ R1 and R3 are in parallel so R13 is 0.5 as you said. R5 and R6 are also in parallel so R56 is 0.5 ohms. R total = R13 + R4 + R56 = 2 ohms, this is sure wrong answers I got here \$\endgroup\$ – Vrabii Daniel Apr 26 '17 at 15:56
  • \$\begingroup\$ So far you are correct \$\endgroup\$ – M.Ferru Apr 26 '17 at 15:57
  • \$\begingroup\$ But the teacher said the correct answer is 0.25 ohms but I seem to get 2 ohms. \$\endgroup\$ – Vrabii Daniel Apr 26 '17 at 15:59
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You AND your teacher are wrong, It helps it you redraw it....

schematic

simulate this circuit – Schematic created using CircuitLab

Notice I added a test voltage and load with meter to each step to verify the changes do not affect the output voltage.

As others have noted, your math for parallel resistors is in error.

It should be \$R_{eq}=R2*R1/(R1+R2)\$, though in this case, where they are the same resistance, it is simply \$R/2\$.

However, I am wondering what happened to R2... So something may be wrong with your original question.

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    \$\begingroup\$ The question is alright, I just named the resistor wrong, indeed it should have been R1, R2, R3, R4 and R5. Thank you for helping. \$\endgroup\$ – Vrabii Daniel Apr 26 '17 at 16:28
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Remember that two resistors in parallel give: Req = R1*R2/(R1+R2)

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I redrew the circuit like so:

schematic

simulate this circuit – Schematic created using CircuitLab

This more clearly shows the actual connections, so you can see at a glance that R1 and R3 are in parallel, for 0.5 Ohm, as are R5 and R6. these pairs are in series, making 1 Ohm. R4 is in parallel with that, for a overall resistance of 0.5 Ohm.

The circuits in this sort of exercise are usually drawn in a way to confuse the student, and to make the thing appear more complicated than it really is. Re-drawing the circuit will usually simplify the problem.

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  • \$\begingroup\$ Thank you. I understand now why 0.5 ohms, redrawing really helps. \$\endgroup\$ – Vrabii Daniel Apr 26 '17 at 16:41

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