Basically, resistance is something that obstructs the flow of charge because of collision of electrons flowing in the conducting wires on either side of the resistor with the ions in the resistor right? So say I have a 9V battery connected to a 100 ohm resistor, the potential drop across it would be the full 9V i.e. 9J of work is done per coulomb of charge.

However, when another resistor is connected in series, say another 100 ohm resistor, the potential drop across the first would (in this case) be halved. How is it possible for less amount of work to be done per unit charge? After all, there is no change in resistance or resistivity of the resistor right?

Have I simply got the definition wrong?

  • $\begingroup$ If your question would be closed, maybe you could give a try to electronics.stackexchange.com , whose topic it better matches. $\endgroup$
    – peterh
    Commented Aug 14, 2016 at 0:34

2 Answers 2


For the case of the $100 \Omega$ resistor connected to the $9 V$ battery means a current of $\frac{9 V}{100 \Omega} = 90 mA$ through the circuit. This means that every second $90 mC$ of charge flows through the resistor and the amount of energy used per second is $9 V \times 90 mC = 0.81 J$. When you have two resistors of $100 \Omega$ is series, the current flowing is $\frac{9 V}{200 \Omega} = 45 mA$ flows through the circuit, which now gives $45 mC$ flowing through the circuit each second giving an amount of energy of $9 V \times 45 mC = 0.405 J$ used. The battery does use $9 J$ of work to push $1 C$ through the circuit, but this amount of charge doesn't flow through the circuit each second.

  • $\begingroup$ This is a circular argument, because the reason we know that two 100 ohm resistors in series adds up to 200 ohms is by Kirchoff's voltage law, which is essentially what OP is asking us to explain. $\endgroup$
    – The Photon
    Commented Aug 13, 2016 at 16:43
  • 1
    $\begingroup$ Do you mean 9V/200Ohms = 45mA? $\endgroup$
    – Ulthran
    Commented Aug 13, 2016 at 16:43
  • $\begingroup$ I think my answer is meant to show that different amount of energies are used in each second in the two different circuits, but the amount of charge of $1 C$ doesn't flow through the circuit in each second. $\endgroup$
    – jim
    Commented Aug 13, 2016 at 16:46

The potential difference between the terminals (eg 9V) determines how much energy will be released by each Coulomb going round the circuit (here 9J/C). Whether you place 1 resistor or 2 or 100 in the circuit, the total amount of energy released per Coulomb flowing through the circuit is the same.

If there are 2 identical resistors instead of 1, the amount of energy per Coulomb released in each will be half of that for the single resistor (4.5J/C instead of 9J/C). This happens even if the 2 resistors are each $1\Omega$ and the 1 resistor is $1000\Omega$!

The current will be different in each case. As expected the power dissipated will be much lower in the case of the $1000\Omega$ resistor. (Power $=V^2/R$.) Power is energy released per second, which is not the same as energy released per Coulomb.


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