# Electrical Resistance, Current and Voltage in Parallel and Series

Why is it that introducing more loads in a parallel circuit decrease equivalent resistance?

While we're on that topic, why is equivalent resistance, a changeable attribute, the resistance of a battery?

And why is the voltage and Current in a series circuit the same across all loads, if the same?

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haruspex
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Why is it that introducing more loads in a parallel circuit decrease equivalent resistance?
Since they're parallel, they all have the same voltage across them. If you introduce more in parallel and the voltage stays the same then there's more total current flowing. So the nett resistance has decreased.
I find it helps to think of it as water flowing in pipes. If you add more pipes in parallel it's easier to push water through.
While we're on that topic, why is equivalent resistance, a changeable attribute, the resistance of a battery?
Is the equivalent resistance the resistance of the battery? First I've heard of it. Can you elaborate?
And why is the voltage and Current in a series circuit the same across all loads, if the same?
The current must be the same, unless charge is being accumulated or dissipated somewhere in the circuit, by a capacitor, say. And if all loads are the same and have the same current then the voltage drops across them must all be the same.

CWatters
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Why is it that introducing more loads in a parallel circuit decrease equivalent resistance?
The waret analogy isn't perfect but... If you connect several pipes to a tank of water (rather than one) it's easier for the water to get out and the flow rate (current) will be higher.

While we're on that topic, why is equivalent resistance, a changeable attribute, the resistance of a battery?
Not sure I understand the question but... You can change the internal resistance of a battery. It can be done by changing the way the battery cells are manufactured. In some cases you can change the internal resistance of a battery by adding more cells in parallel.

And why is the voltage and Current in a series circuit the same across all loads, if the same?
What haruspex said. Imagine a long hose with various valves or restrictions in it it....

Current - Water isn't compressible and if there are no leaks all the water that goes in one end must come out the other. eg same current flows through all components.

Voltage - Each restriction will cause a small pressure (=voltage) drop. Add them up and the total will equal the pressure drop between the two ends.

Don't get too fixated on the water model though because it isn't perfect.

To clarify,

By using a parallel circuit, more branches means each load has a higher voltage input, thus allowing a higher value for V in R=V/I. But that would mean parallel reduces resistance. What I'm asking is why if I have a parallel circuit and I fit another five light bulbs to one of the circuits, the equivalent resistance equations report a lower Re than if I didn't have those five light bulbs. This doesn't seem to fit with the pipe analogy.

My science teacher says that Re is the internal resistance of a battery. Internal resistance seems to be something built into the battery. However, how can internal resistance equal Re? And a battery "recycles" electrons so as the source of voltage and current, how can it reduce current?

If I have 2 light bulbs in series, an ammeter will say the same current wherever put it. Wouldn't the current be greater before the first load that after the second? And if I have thirty light bulbs (more than a battery can sustain) in series, why aren't the first few brighter than the last?

CWatters
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You seem to be confusing a lot of different things.

By using a parallel circuit, more branches means each load has a higher voltage input, thus allowing a higher value for V in R=V/I.
That doesn't make any sense. If you add loads in parallel with existing loads then all the loads will have the same voltage. .

What I'm asking is why if I have a parallel circuit and I fit another five light bulbs to one of the circuits, the equivalent resistance equations report a lower Re.
What do you mean by "fit another five light bulbs to one of the circuits"?? Do you mean fit in parallel with or in series with? It makes all the difference. Best draw a before and after circuit so we know what you mean.

My science teacher says that Re is the internal resistance of a battery.
That's another subject. The notation Re can mean anything you want. . Re could mean "the equivalent resistance of the battery" OR "the equivalent resistance of part of the circuit we discussed last wednesday". Context is everything. In maths "X" and "y" get reused a lot. It's the same with electronics.

In the first part above you talk about Re being the equivalent resistance of several loads. You can use Re to refer to the internal resistance of a battery BUT best not to do that if you are talking about the same circuit or it will get confusing.

Lets call the internal resistance of the battery Rint from now on...

Internal resistance seems to be something built into the battery. However, how can internal resistance equal Re? And a battery "recycles" electrons so as the source of voltage and current, how can it reduce current?
Batteries are not ideal voltage sources. If you were to try and draw a lot of current from a AAA battery it would not be able to maintain 1.5V. This effect can be modeled by treating the cell as an ideal voltage source in series with a resistor Rint.

It's Rint+Rload because the load is in series with the internal resistance.

You can see that if Rint is increased the current will fall. As will the voltage measured at the battery terminals.

If I have 2 light bulbs in series, an ammeter will say the same current wherever put it. Wouldn't the current be greater before the first load that after the second?
No. If the current wasn't the same you would have more current arriving at a node than leaving it. That would be like a water pipe with more going in than coming out.

And if I have thirty light bulbs (more than a battery can sustain) in series, why aren't the first few brighter than the last?
Given that the same current flows through each (see above) AND each bulb has the same resistance then the voltage drop across each bulb will be the same V=IR.

The power in each bulb is P=IV. If I and V are the same then the power and brightness is the same.

You seem to be confusing a lot of different things.
That doesn't make any sense. If you add loads in parallel with existing loads then all the loads will have the same voltage. .
Why won't the first few loads have higher voltages? How is Vsource=V1=V2 possble?

What do you mean by "fit another five light bulbs to one of the circuits"?? Do you mean fit in parallel with or in series with? It makes all the difference. Best draw a before and after circuit so we know what you mean.
Sorry. I meant into one of the branches of the parallel circuit. I'm just asking why more loads in parallel circuits reduce Re of the whole circuit.

That's another subject. The notation Re can mean anything you want.
In this case, the Re of the whole circuit.

Batteries are not ideal voltage sources. If you were to try and draw a lot of current from a AAA battery it would not be able to maintain 1.5V. This effect can be modeled by treating the cell as an ideal voltage source in series with a resistor Rint.
So you're saying batteries create resisitance because they are inefficient?

No. If the current wasn't the same you would have more current arriving at a node than leaving it. That would be like a water pipe with more going in than coming out.
So in a circuit with more loads are you narrowing this water pipe to reduce current?

The power in each bulb is P=IV. If I and V are the same then the power and brightness is the same
Why wouldn't V be reduced as it gets used by the first few loads?

CWatters
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Perhaps best to take things one step at a time.

Lets say you have two circuit nodes A and B. The voltage difference between them is VA - VB.

If you connect a resistor between those nodes the voltage drop across the resistor will also be VA - VB.

Two components are considered to be in parallel if they are connected between the same two nodes. So a second resistor connected in parallel to the first must also be connected to nodes A and B. So the voltage drop across both resistors will be VA - VB.

CWatters
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I'm just asking why more loads in parallel circuits reduce Re of the whole circuit.
Lets say you have resistor R1 connected between nodes A and B. The current flowing in R1 will be given by Ohms law...

I1 = V/R1 ........................(1)

Now connect another resistor R2 in parallel. The current flowing in R2 will be..

I2 = V/R2 .........................(2)

So the total current Itotal flowing between A and B will be..

Itotal = I1 + I2

substitute from (1) and (2)

Itotal = V/R1 + V/R2

The equivalent resistance Re = V/Itotal

= V/(V/R1 + V/R2)

= 1/(1/R1 + 1/R2)

That will allways be smaller than R1 or R2

Well, this all makes sense theoretically but the fact that by adding more barriers to the current I help it flow makes no sense to me.

CWatters