# Potential Difference across resistors

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?

• If your question would be closed, maybe you could give a try to electronics.stackexchange.com , whose topic it better matches. – peterh Aug 14 '16 at 0:34

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.
• 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. – jim Aug 13 '16 at 16:46
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.