In my physics class, we did an experiment in which we supported Ohm's law. We used a simple circuit that includes one resistor. The relationship of I = VR, and variants thereof, became apparent. We mainly looked at the relationship between potential difference and the current in series.

What we have not done is see if this relationship is apparent in a series circuit with multiple resistors. I've did further research and read about Non-ohmic resistors, do they have anything to do with this? I can't really make the connection.

So my question is, what is the relationship between the potential difference across two resistors in series and the current through them? We are moving towards parallel, but I would like to know this question better.


Two resistors in series will have the same current flowing through them.

The potential difference across each resistor ($V=IR$) will sum together.

So, for example if you connect a battery with a given voltage $V$ to the two resistors ($R_1$ and $R_2$), the voltage drop across the resistors will equal the battery voltage, and then you will know the current from $V = I(R_1+R_2)$).

  • $\begingroup$ So basically, V is still equal to IR, R being the sum of the two resistors? What about the voltage across them? And the current across them? $\endgroup$
    – Joel Aqu.
    May 10 '14 at 10:51
  • $\begingroup$ Yeah. The current depends on V as above. The voltage of the first resistor V1 = R1/(R1+R2)*V, and V2 = R2/(R1+R2)*V $\endgroup$
    – xyz
    May 10 '14 at 11:15

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