# Why is the current the same in series circuits?

erocored
If the current goes through the first end of the resistor will it be less than on the other end of resistor?

Mentor
If the current goes through the first end of the resistor will it be less than on the other end of resistor?
No. Charge is conserved so that would lead to enormous charge accumulation in the resistor.

etotheipi, vanhees71 and erocored
2022 Award
If the current goes through the first end of the resistor will it be less than on the other end of resistor?
Current is the amount of charge passing a point in unit time. That means it's the number of electrons passing a point in unit time multiplied by the charge on an electron. So if the current is different electrons must be stopping somewhere and accumulating - which would lead to a charge buildup which we don't see in practice. So the current must be the same.

etotheipi, berkeman, vanhees71 and 1 other person
Mentor
If the current goes through the first end of the resistor will it be less than on the other end of resistor?
Just to add to the other replies, the Current is the same through the resistor, but the Voltage drops as the current flows through the resistor. The Voltage drop across a resistor as a Current I flows through it is given by the famous Ohm's Law equation: ##V = IR##.

https://en.wikipedia.org/wiki/Ohm's_law

etotheipi
If you want the fancy-schmancy version of what was said above, you can construct a (mathematical) closed surface ##\Sigma## around the entire resistor, and then apply the continuity equation ##\frac{\partial \rho}{\partial t} + \nabla \cdot \vec{j} = 0##, i.e.$$\dot{Q} = -\int_{\Sigma} \vec{j} \cdot d\vec{S} = I_{\text{in}} - I_{\text{out}}$$In a steady state ##Q = \text{constant}## within that closed surface, so ##\dot{Q} = 0## and ##I_{\text{in}} = I_{\text{out}}##

Delta2, weirdoguy and berkeman
Gold Member
If you want the fancy-schmancy version of what was said above, you can construct a (mathematical) closed surface ##\Sigma## around the entire resistor, and then apply the continuity equation ##\frac{\partial \rho}{\partial t} + \nabla \cdot \vec{j} = 0##, i.e.$$\dot{Q} = -\int_{\Sigma} \vec{j} \cdot d\vec{S} = I_{\text{in}} - I_{\text{out}}$$In a steady state ##Q = \text{constant}## within that closed surface, so ##\dot{Q} = 0## and ##I_{\text{in}} = I_{\text{out}}##
Don't worry if you don't understand this yet. It isn't necessary for students learning basic electronics. You'll get to this later, if you keep studying physics/electronics.

erocored, Delta2 and berkeman
Gold Member
2022 Award
One should, perhaps, also mention that all this is correct only approximately. It is exactly valid for DC circuits and approximately for AC circuits if the geometric extension ##d## of the entire setup is small compared to the wavelength of the electromagnetic radiation associated with the frequency of the AC, i.e., for ##d \ll \lambda## with ##\lambda=2 \pi c/\omega=c/f##.

If you have very long wires, then the electromagnetic fields as well as the currents are wavelike propagating along these wires. See, e.g.,

https://en.wikipedia.org/wiki/Telegrapher's_equations

Delta2
Gold Member
If the current goes through the first end of the resistor will it be less than on the other end of resistor?
If that were true then where would all those surplus electrons go? (I do love a noddy answer then one is available.)

Homework Helper
If that were true then where would all those surplus electrons go? (I do love a noddy answer then one is available.)
They could, of course, pile up in the resistor. But like charges repel. If the electrons build up in the resistor, only so many can be added before the repulsion becomes too significant and fewer start arriving.

Indeed, this happens. It is just that the force of repulsion between electrons is quite large. The repulsion gets huge before the electron build-up becomes big enough to notice.

How big is the repulsion? Let us work it out on the back of an envelope.

Suppose that we managed to build up one coulomb of charge in a resistor that is 1 cm in size. For ease of calculation, let us make this resistor spherical and makes its radius 1 cm. The charge will distribute itself on the outside of the sphere. This is a consequence of Newton's shell theorem.

We want to calculate the potential of this sphere. How many volts does it take to push additional current into this shell from far away? A Volt is a Joule per coulomb. So asking the question another way, what we want to know how many Joules it would take to move a unit charge from infinity to the spherical shell.

$$F=\frac{kq_1q_2}{r^2}$$
Here, ##F## is the electrostatic repulsion. ##q_1## is the charge on the sphere, ##q_2## is the amount of charge we are trying to bring in and ##r## is the current distance between our charge and the spherical resistor.

As we bring the test charge in, the repulsive force ##F## will oppose our efforts, of course. In first year physics, if we were doing work against a force, we would multiply force times distance. Since the force is variable here, we have to do an integral.
$$W=\int _\infty ^r \frac{kq_1q_2}{x^2} dx$$
We are starting at ##x=\infty## and ending at ##x=r## and adding up the incremental work ##F \cdot dx## done over all the tiny bits of path (##dx##) in between.

This is an easy integral. The result is well known to be $$W=\frac{kq_1q_2}r$$
[This is back of the envelope work -- I am not being careful about signs]

We are after a figure for work done per charge transported. That is, we want:$$\frac{W}{q_2}=\frac{kq_1}r$$

Now let's put in the givens of the problem. We are working in SI so ##k## is approximately 9 x 109 kg meter3/sec2/coulomb2. ##r## is 0.01 meter and ##q_1## is 1 coulomb.

So we get:$$\frac{W}{q_2}=\frac{9 \times 10^9}{0.01}=9 \times 10^{11}$$
The units all work out so that this is 900 billion Joules per coulomb. [For reference, that's about 200 kilotons per coulomb]

900 billion volts.

Or, put another way, a C cell can succeed in cramming one or two trillionths of a coulomb of excess charge onto a 1 cm resistor. It could do this with a one microamp charge rate in a microsecond or two.

Last edited:
sophiecentaur, erocored, Merlin3189 and 1 other person
It may be helpful to compare simple electrical circuits with pipes carrying water. (Helpful, but only up to a point!)

For example:

If the current goes through the first end of the resistor will it be less than on the other end of resistor?

Here we can think of a resistor as a piece of very thin pipe that water can't flow through very easily. A wire is a thick pipe. So you have a thick pipe (wire) feeding water to a thin pipe (resistor) and then at the other end of the thin pipe, we have a thick pipe taking water away.

Would you expect to see less water flowing out of the thin pipe than what flowed in? Normally, you wouldn't! And you can't expect to see more water leaving the output end either.

The exception is if the thin pipe is leaking, OR if the thin pipe is inflating like a balloon. In these cases the water flowing out would be less than the water flowing in. But if the pipe is behaving normally, then the flow out should be equal to the flow in.

The balloon case is interesting, because it's a bit like a capacitor - a storage or accumulating element. In high frequency circuits (radio frequency circuits, for example), it's possible for a resistor to behave like a balloon / capacitor and mess up the circuit operation -- so you can't use normal resistors in those applications.

erocored