# Basic question regarding resistance

Thread Starter

#### cm101

Joined Dec 26, 2011
3
Hi guys. I'm new here and need a little help with a very beginner question about how a resistor would affect the current in a circuit.

Basically, if I had a simple circuit consisting of just a battery and a wire going from the + to - terminal, and I then placed a resistor in this circuit, would the resisitor cause the current to decrease at every point in the circuit or just across the resistor itself?

I know this is probably an easy one to answer, but I can't get my head around it.

#### evilclem

Joined Dec 20, 2011
116
Resistance in a circuit like this can be thought of like a hose full of water. If you put a bit of small hose in the middle it will restrict the flow of water through the entire hose. Thus a resistor affects the current flowing through the whole circuit.

Thread Starter

#### cm101

Joined Dec 26, 2011
3
Thanks for the quick reply. I thought this would be the answer and this is why I was getting confused.

Would the electric field strength that causes the electrons to move still not be the same right through the circuit (i.e. we haven't changed the battery voltage, just added a resistor)? So my thinking is that (in comparison to the circuit without the resistor), the electric field is still moving the exact same number of electrons as before in the wire part of the circuit. However, this same field struggles to move as many electrons is the resistor so the current decreases here but remains the same everywhere else.

I hope i've explained that clearly! I can see why the answer you gave is the correct one, but my thinking about how the electric field works is causing me confusion.

#### Wendy

Joined Mar 24, 2008
22,436
• jaygatsby

#### Adjuster

Joined Dec 26, 2010
2,148
Thanks for the quick reply. I thought this would be the answer and this is why I was getting confused.

Would the electric field strength that causes the electrons to move still not be the same right through the circuit (i.e. we haven't changed the battery voltage, just added a resistor)? So my thinking is that (in comparison to the circuit without the resistor), the electric field is still moving the exact same number of electrons as before in the wire part of the circuit. However, this same field struggles to move as many electrons is the resistor so the current decreases here but remains the same everywhere else.

I hope i've explained that clearly! I can see why the answer you gave is the correct one, but my thinking about how the electric field works is causing me confusion.
Hello CM101, welcome to the forum. If you want to learn about electronics, it is good to get the basics sorted out first. This will save a lot of trouble later on, so you are trying to do the right thing. Here are a few things to think about.

A simple loop of components with nothing branching off it is called a series circuit. This is what you are describing with a battery and a resistor. The current at any point in a series circuit is the same.

Think about it: how could it be otherwise - if the currents through the battery and the resistor were different, there would have to be an accumulation of electric charge at the point where they join. For a DC circuit, this would imply charge accumulating indefinitely: this cannot happen. Later on you may learn about this formally in the form of Kirchhoff's current law.

You also need to think how the electric field is distributed in the circuit: it is not the same when the resistor is added. With no resistor, all* the voltage is applied along the length of the wire, so the electric field lies along it, driving a large current through the wire's very low resistance. When the resistor is added, some of the voltage will be dropped across it, leaving less voltage along the length of the wire. This is a result of Kirchhoff's voltage law.

http://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws

If you imagine the circuit sliced into small elements along its length, the electric field strength is therefore not the same everywhere, but at any point it will be at just the level required to drive the same current through the element concerned.

*When a battery is short-circuited like this, quite a lot of the voltage may be lost in resistance inside the battery - but don't worry if you have not dealt with internal resistance yet.

#### JDT

Joined Feb 12, 2009
657
Basically, if I had a simple circuit consisting of just a battery and a wire going from the + to - terminal, and I then placed a resistor in this circuit, would the resisitor cause the current to decrease at every point in the circuit or just across the resistor itself?
In a series circuit, the current is the same at every point in the circuit.

Sounds simple - and it is - but this is a fundamental truth in circuit analysis.

Your series circuit consists of the resistor, the wire (which also has resistance) and the battery (which has internal resistance as well as a source of EMF in series).

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Thread Starter

#### cm101

Joined Dec 26, 2011
3
Have you read the AAC book and it's take on resistance?

http://www.allaboutcircuits.com/vol_1/chpt_1/5.html
Yep, thanks. It is a good read, but there are probably some gaps in my knowledge which is why I am not understanding it properly.

You also need to think how the electric field is distributed in the circuit: it is not the same when the resistor is added. With no resistor, all* the voltage is applied along the length of the wire, so the electric field lies along it, driving a large current through the wire's very low resistance. When the resistor is added, some of the voltage will be dropped across it, leaving less voltage along the length of the wire. This is a result of Kirchhoff's voltage law.

http://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws

If you imagine the circuit sliced into small elements along its length, the electric field strength is therefore not the same everywhere, but at any point it will be at just the level required to drive the same current through the element concerned.

*When a battery is short-circuited like this, quite a lot of the voltage may be lost in resistance inside the battery - but don't worry if you have not dealt with internal resistance yet.
Thanks for the explanation. I think I will do some more reading regarding electric fields before I ask any more questions as I think that is where my understanding breaks down at the moment.

#### Adjuster

Joined Dec 26, 2010
2,148
One of the most basic relationships in electronics is Ohm's Law, which tells us that the current through (ordinary) conductors is proportional to the voltage applied across them. One result of this is that in a simple circuit where a battery voltage of V volts is applied across a resistance of R ohms, the current of I Amps is given by I = V/R.

The voltage dropped along the wiring to the resistor is usually negligible in comparison to what appears across the resistor, but if you could measure it, it would be just enough to explain the current flowing through it.

Thus the electric fields in a series circuit are so distributed as to agree with the current being equal throughout it.

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