I have a serial and a parallel circuit. In both of the circuits I am using two resistors.These two resistors have the same resistance $a$.

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In both the circuits I have tried reading the amount of current with different amount of voltage's. I've gotten the following results.

$$ \begin{array}{c|lcr} \text{Voltage (V)} & \text{Serial (mA)} & \text{Parallel (mA)} \\ \hline 1 & 0.42 & 2.91 \\ 2 & 0.98 & 4.83 \\ 3 & 1.47 & 6.8 \\ 4 & 2.03 & 8.96\\ 5 & 2.65 & 11.17 \\ 6 & 3.27 & 13.05 \\ 7 & 3.72 & 15.37 \\ 8 & 4.28 & 17.13 \\ 9 & 4.84 & 19.13 \\ 10 & 5.31 & 21.7 \end{array} $$

The total resistance in the serial circuit would be $2a$. For the parallel circuit it would be $\frac{a}{2}$.

I then use the formula $V = IR$ to find the resistance for each different voltage I used and current I read on the currentmeter (check the table).

As it turns out, the average amount of resistance in the serial circuit is: $1959$ Ohms. In the parallel it's $440$ Ohms.

$2a = 1959 \iff a = 979.5$

$\frac{2}{a} = 440 \iff a = 0.0045454545...$

Why is the resistance different in the parallel circuit opposed to the serial circuit?

Are my readings incorrect? Are my calculations incorrect? I'm very new to physics.

  • $\begingroup$ You correctly concluded that the resistance of the parallel circuit would be $\frac{a}{2}$, but then you went on to write $\frac{2}{a} = 440$. You maybe would have spotted that mistake if you had used units properly; if $a$ is a resistance, it has units of Ohm, written $\Omega$ (or some other equivalent unit), so that your equation would have read $\frac{2}{a}=440\Omega$, leading you to $a\approx0.0045\Omega^{-1}$, which clearly is not a resistance. To answer your question regarding your readings, it would be helpful if you described how you connected your amperemeter to the circuit. $\endgroup$
    – RQM
    Commented Feb 13, 2015 at 14:33
  • $\begingroup$ The resistance in the serial configuration should be 4 times bigger than in the parallel. Right? Then the current should be, in each case and case 4 times smaller. Right? But it is more than 4 times smaller. So, there is some more resistance in the serial configuration. I don't know where from. (The wires themselves?) Now, the resistance in parallel is not $2/a$ but $a/2$. $\endgroup$
    – Sofia
    Commented Feb 13, 2015 at 14:41
  • $\begingroup$ @Sofia and RQM Ah yes. I made a mistake there. It seems that with parallel it's 880 then. Which seems to relate to 979.5. I think the meter I used at school isn't very precise, which explains why theres a pretty big difference between 880 and 979.5. I actually used a multimeter and connected it serially like the teacher told me. Thanks for the hints guys, this should suffice. At least I understand the theory behind it correctly. $\endgroup$ Commented Feb 13, 2015 at 15:17

1 Answer 1


$\frac{2}{a} = 440$

It should rather be

$\frac{a}{2} = 440$

So you get $a=880$ for the parallel ciruit. There is still some difference between the two measured resistances. It looks like your ampere meter has an internal resistance of 66 Ohm or your volt meter has an internal resistance of 13 kOhm:

If the ampere meter is in series with the volt meter, you will measure a higher current than expected, because some current will go through the volt meter. If on the other hand your volt meter is parallel to the ampere meter, you will get higher than expected voltage readings, because there is a potential difference at the ampere meter. Update: this does not explain the different resistances, because you would expect the opposite direction of the effect.


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