# Potential diff

Hi,

When you put different resistors, lets say some light bulbs, in a circuit in parallel, how come there can exist a greater potential difference over the light bulbs in parallel then in series with the same light bulb. I know this is because there is a greater current, but how come it affects the potential difference? There is still the same energy/coulomb needed to pass through the resistor, no? So how is the higher current, the higher amount of charged particles, affecting the amount of energy(pot. diff.) needed to pass through the resistor? Are the charges between themselves affecting the energy needed?

My second question is: how come you can you put a higher watt light bulb in a circuit, which has subsequently more resistance, and still create the needed amps? If you try this with a simple parallel circuit, the higher resistance you create the less amps you will get.
So how come a light bulb or a washing machine can get the needed amps? The washing machine probably uses a transformator right?, but what about the light bulb then?

Anyways, thnx in advance to anyone who likes to help me out a bit.

## Answers and Replies

Homework Helper
To your first question, can you clarify something? Are you talking about the potential differnce across the entire parallel segment vs the total potential difference across both bulbs in series (not just each bulb separately)?

If so, then this has to do with the internal resistance of the battery (or other voltage source). Since more current is drawn throught he circuit, the internal resistance (little "r" usually) will cause a voltage drop that is proportional to the current. This means that some of the potential difference of the battery (the original "emf") is used up inside the battery (V=Ir) causing the remaining potential differnce to be less than emf.

This is why, inexplicably to some teachers, bulbs connected in parallel keep getting dimmer with each bulb added, contrary to what they just taught.

And about the second question is: a higher watt light bulb has less resistance than a lower watt light bulb.
A "60 watt" bulb for your house is only a 60 watt bulb if it is put in 120 volt circuit.

THat which is "set" for any appliance is its resistance (or impedence for the washing machine). The appliance will get the current that is determined by Ohm's Law.

(Be aware that the resistance of a light bulb increases very significantly as it heats up, so the cold resistance of a bulb is nothing compared to its working resistance)

thnx

1q: So you mean that the energy used by the bulb (pot. diff. over the bulb) in parallel is "actually" still the same as in series but we measure the full, lets say 12v (battery), 12v pot. diff. because some of the enrgy/coulomb(pot. diff.) is used up because the battery is creating a resistance proportional to the amount of current?

2q: Does that mean that a higher watt applience(in respect to 120 v) usually has less resistance? Or when it hasn't it uses a transformator.

Thnx.

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1q: what I mean is how come there can be a greater pot. diff. over a single bulb in parallel(using other bulbs) then is series? Where does the higher potential drop come from over the bulb? The potential drop over the bulb should stay the same, no, there's the same obstacle(resistance) to overcome, no? just more amps flowing across it. If the potential drop comes from the extra resistance created by the current in the battery, then it makes -almost- sence, but a voltmeter should not measure the extra drop over the battery but only the pot. drop over the bulb?

Mentor
Voltage drop depends on the makeup of the entire circuit, but the total voltage drop is equal to the voltage of the source. The only property inherrent to the light bulb (resistor) is resistance.

Staff Emeritus
Gold Member
When you connect resistors in parallel, you are applying the same potential difference to each resistor. Look at your wires -- (ideal) wires are by definition at the same potential all along their lengths, since their (ideal) resistance is zero ohms. Notice that the wire leaves the battery's positive terminal and is connected directly to one side of each resistor on the "top" of the circuit. Therefore, all of the resistors experience the same potential on that side. The same argument applies to the bottom side of the circuit, also -- so all the resistors experience the same potential difference.

I believe you may be asking something like this: "why doesn't the potential difference experienced by the resistors decrease as you look further out in the 'ladder?' Why doesn't the light bulb nearest the battery light up more than the one far away?"

The answer is in that condition of ideality I mentioned a minute ago -- wires are usually considered to have zero resistance at DC for short lengths -- not just low resistance, but ZERO. If this is true, then you can see how all points on the wire have the exact same potential. On the other hand, ideal wires don't exist. All wires have some resistance. So, in the end, you are correct -- every resistor in a parallel chain does NOT have the same potential difference applied to it in a real circuit. The light-bulb furthest from the battery DOES experience a slightly lower potential difference, and isn't QUITE as bright as the one nearest. However, the resistance of a real wire is just about completely insignificant compared to the resistance of a light-bulb -- we're talking something like five orders of magnitude difference. As a result, it's much simpler (thought not absolutely correct) to treat the wire as having zero resistance.

Try to remember the following fact: there is only a voltage drop where there is resistance.

In a series circuit, each light-bulb has resistance, and thus each light-bulb demands a potential difference across it as per Ohm's law. Think about two light bulbs wired in series -- each one experiences half the potential difference of the battery. Why? Continuity. Start at the battery's positive terminal. Say it's +12V. Now move your finger through the first light bulb. Say to yourself "that's a voltage drop of 6V." Now move through finger through the second bulb. Say to yourself "that's another six volts. That means this point in the circuit, at the bottom, should be 0V." Sure enough, the bottom of the second bulb is connected via an (ideal) wire to the negative terminal of the battery, and is thus at 0V. You can follow any loop through a circuit, and continuity (and common sense) requires that traversing a loop always gets you back to the starting voltage. This is known as Kirchkoff's Voltage Law, if you want to put a name to it.

- Warren

Staff Emeritus
Gold Member
Originally posted by russ_watters
Voltage drop depends on the makeup of the entire circuit, but the total voltage drop is equal to the voltage of the source. The only property inherent to the light bulb (resistor) is resistance.

To expand on this. First look at http://home.comcast.net/~rossgr1/Math/circiut.pdf drawing. Observe that the circuit on the left labeled SERIES has only a single path while the circuit on the right, labeled PARALLEL, has 2 separate paths. Current flow is the physical movement of electrons, like water they will follow all available paths.

Are you familiar with Ohms Law?

V= IR

Where V= voltage, I= current, R = Resistance.

This relationship holds for the entire circuit and each circuit element.

Here are some simple circuit rules.
In a series circuit:
1. the current is constant everywhere.
2. The voltage drops across each resistance sum to the source voltage.
3. The total resistance is the sum of all circuit resistances.
In a Parallel circuit:
1. The current at the sources is the sum of the current in each branch.
2. The Voltage of each branch is equal to the source voltage.
3. The total resistance is the inverse of the sum of the inverses of the branch resistance.

the 3rd statement in each set above is really a restatement of the first 2. That is it can be derived given 1 and 2 as follows.

Give that the current is constant everywhere and that the voltages sum to the source we can write.

VT=V1+V2
Then from Ohms law
VT=IRT
V1=IR1
V2=IR2

Combining these gives

IRT=IR1+IR2
now cancel the constant current from each term.

RT=R1+R2

For the Parallel circuit we have a constant voltage and

IT=I1+I2
Once again from Ohm's Law
IT=V/RT
I1=V/R1
I2=V/R2
Now combine the relationships to get

V/RT=V/R1+V/R2
Now cancel the constant current to get the result
1/RT=1/R1+1/R2

If you play a bit with the expression for a parallel resistance you will see that the total resistance is LESS then the smallest resistor.

I have presented a quite a bit of stuff here, so look at this and think about your light bulbs. See if any bulbs light up!

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Staff Emeritus
Gold Member
As per your second question -- a "higher wattage" light bulb has LOWER resistance.

A 120W light bulb on 120V will carry 1A of current. 120V * 1A = 120W. Ohm's law says it has a resistance R = V/I = 120 ohms.

A 60W light bulb on 120V will carry 0.5A of current. 120V * 0.5A = 60W. Ohm's law says it has a resistance R = V/I = 240 ohms.

- Warren

Homework Helper
OK, your question is more basic than my previos answer. I don't think you need to worry about internal resistance yet.

When the current goes through a single path, the total potential must "drop" the full potential difference as it goes the two bulbs. Each unit of charge has to flow through each bulb. Each electron starts with "so much" energy. In a single pathway, or series circuit, the energy per charge (that's potential) is split between the bulbs.

If there are two separate pathways, then each electron chooses a path. One or the other, but each electron still has the full energy per charge (that's potential) to devote to a single bulb.

Any better?

edit: wow a bunch of people posted whilst I wrote this.

thnx

Thanx so much, I wasn't actually certain why in some cases there could be devoted more energy to one certain light bulb with the same amount of resistance.

Thnx for enlightening me guys.

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So, would it be valid to state: The larger the voltage drop across a resistor, the smaller the current flowing through it? Or am I really confused?

Staff Emeritus
Gold Member
Originally posted by FestiveF
So, would it be valid to state: The larger the voltage drop across a resistor, the smaller the current flowing through it? Or am I really confused?

You are exactly backwards, the larger the voltage drop the higer the current.

E = IR

E is voltage,I is current, R is resistance.

If the resistance is constant, a higher voltage must result in a higher current.