# How to calculate the resistance of a series circuit?

• erocored
In summary, current is the electrons that move due to potential difference. When electrons pass through a resistor they lose some of their kinetic energy. This causes the current to be the same as the resistance.

#### erocored

Why we say that the resistance of the series curcuit is equal to the sum of the resistances of the resistors?

It's not equal, it's equivalent. This means that if you replace the two resistors with a single resistor, it will draw the same current.

erocored
kuruman said:
It's not equal, it's equivalent. This means if you replace the two resistors with a single resistor, it will draw the same current.
Why is it possible?

erocored said:
Why is it possible?
Ohm's law makes it possible. Say you connect a ##5 \mathrm{\Omega}## resistor to a 10 V battery. Ohm's law, ##V=IR## says that the current is ##I=V/R=2~ \mathrm{A}##. Now connect two resistors, ##R_1=2 \mathrm{\Omega}## and ##R_2=3 \mathrm{\Omega}##, in series to the same battery. "In series" means the current through each resistor is the same. Call that current ##I'##. The voltage across the series combination is the voltage across the battery, 10 V. It is also the sum of the voltage drops across each resistor. Then $$I'R_1+I'R_2=I'(R_1+R_2)=I' (5\mathrm{\Omega})=10~\mathrm{V}.$$You can see that the combination draws the same current, i.e. ##I'=I=2~ \mathrm{A}.##

@erocored, In this thread and your other thread, you seem to have a misconception about what electric current is. Your questions have been answered, but that doesn't seem to help. Tell us in your own words what you think current is, and how you think it should behave.

russ_watters
anorlunda said:
@erocored, In this thread and your other thread, you seem to have a misconception about what electric current is. Your questions have been answered, but that doesn't seem to help. Tell us in your own words what you think current is, and how you think it should behave.
I think the current is the electrons that move because of potential difference. When electrons passing through a resistor they lose some of their kinetic energy. I guess that we can calculate common resistance by the sum of the resistance R1 and R2 because total less of energy wil be equal to loss of energy if current goes through the resistor of resistance R1+R2. But I can't understand why the electrons don't get energy going through BC?

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vanhees71
Your explanation is almost correct. You can imagine the electrons as moving at constant speed. They do not lose kinetic energy. They lose potential energy when the go through a resistor. The potential energy lost is converted into heat. The rate at which energy is lost in the resistor (power) is given by ##P=I^2R##. Segment BC in your drawing has zero resistance, therefore no power is lost there.

vanhees71, erocored and etotheipi
erocored said:
When electrons passing through a resistor they lose some of their kinetic energy.

That is part of your wrong idea. The kinetic energy of electrons has almost nothing to do with electric energy. Even though voltage changes move through wires at close to the speed of light, the drift velocity of electrons is usually less than 1 cm/second.

Electric power, which is the rate of delivery of electric energy, is proportional to voltage times current. In a superconductor, you can have zero voltage, nonzero current, and zero power lost in that section.

The actual electric energy is carried in the electromagnetic fields, not in kinetic energy of the electrons. It is rather advanced to explain how that works.

But the best thing you can do to understand basic electricity is to forget that you ever heard of electrons. Think of voltage V, current I, resistance R, and power P. V=RI, and P=VI. Those are what you need to analyze a circuit at the B level.

To fully understand what happens inside a wire, you first need to study Maxwells Equations, then the Drude Model, and then the free-electron model. Those are I or A level topics.

erocored and DaveE
erocored said:
But I can't understand why the electrons don't get energy going through BC?
Because there are no power sources between B and C.

Imagine a ball is pushed into a tube at the negative side of the battery, such that the ball attains a certain velocity. Then, at A, the ball is slowed by R1. Once at B, it reaches a new (slower) velocity and continues to roll at that velocity until it reaches C. Then it is slowed down again through R2, having a new velocity ##v_f## at D. And the ball will keep rolling down the tube at ##v_f## until it reaches the positive side of the battery.

Between B and C, there are no gains or losses in energy. Just like between the battery and A or between D and the battery.

erocored
I think it's also good to have a more accurate microscopic picture about the resistor. Then the confusion about kinetic energy and potential energy lost is easily resolved.

If you have a piece of metal there are conduction electrons which can move easily within this metal, but due to perturbations of the ideal crystal lattice (among other things also the thermal motion of the lattice are such perturbations) there is friction. The equation of motion of an electron along a wire (which we take along the ##x##-direction) thus is
$$m \ddot{x}=-e E - m \gamma \dot{x},$$
where ##E## is the electric field along the wire and ##F_{\text{fr}}=-m\gamma \dot{x}## is the friction force taken to be linearly dependent on the velocity, which is justified, because the velocity of an electron in normal house-hold wires is very very slow (at the order of 1 mm per second).

Now if you have a DC current, you have ##\dot{x}=v=\text{const}## and thus ##\ddot{x}=0##. This implies
$$m \gamma v=-e E.$$
If now ##n## is the conduction-electron density the current through the wire is
$$j=-n e v=-n e \frac{-e E}{m \gamma}=\frac{n e^2}{\gamma} E.$$
This means that the electric conductivity is
$$\sigma=\frac{n e^2}{\gamma}.$$
The energy loss per unit volume and per unit time is
$$\frac{P}{V}=j E = \frac{j^2}{\sigma}.$$
If the total length of the wire is ##l## and its crossectional area ##A## you get the power ( Ohmic energy loss per unit time)
$$P=\frac{j^2 A l}{\sigma} = \frac{l}{A \sigma} I^2 \; \Rightarrow\; R=\frac{l}{A \sigma}.$$
so the energy loss is due to friction of the conduction electrons within the wire. That's the Drude model of DC electric conductivity/resistance.

Delta2, etotheipi, Lnewqban and 1 other person
erocored said:
. When electrons passing through a resistor they lose some of their kinetic energy.
Without actually doing any sums, just consider how much kinetic energy the electrons could be carrying. The total mass of the 'mobile' electrons in a conductor is a tiny fraction of the conductor's total mass {1/(1800Xatomic mass} with just one electron from each atom) and then that the average drift speed of electrons is about 1mm/s. Yet that current of electrons can be transferring kilowatts of power. What counts is the Potential Energy that's lost on the way through the device and there always IS a voltage (Potential) drop on the way through. If you measure the VOLTS between the negative terminal of the supply and various points around the circuit, it gets less as you go round, until you reach the negative terminal itself, where there is 0V across your meter.
The Energy Lost can either be through heating or by work done by a motor that's connected.

erocored
Well, the kinetic energy of the conduction electrons is negligible in energy transport for usual household currents. What transports the energy along the circuit is the electromagnetic field. The best treatment of this in my opinion is found in Sommerfeld, Lectures on Theoretical Physics, vol. 3, where he treats the DC coax cable in detail (in the fully sufficient non-relativistic approximation).

As I tried to show above with the Drude model the resistance (or equivalently the electric conductivity) is a transport coefficient related to the friction of the conduction electrons in the metal.

etotheipi, Delta2, davenn and 2 others
vanhees71 said:
Well, the kinetic energy of the conduction electrons is negligible in energy transport for usual household currents. What transports the energy along the circuit is the electromagnetic field.

That is it in a nutshell

sophiecentaur