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Resistor Voltage Drop Concept Question 
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#1
Feb1412, 06:43 PM

P: 35

I'm currently studying electric circuits and can't find the answer to my question anywhere so far. I realize current is the flow of positive charge even though positive charges stay put in reality and that this is an old historic concept that is still around today because Benjamin Franklin designated positive charge as the one that moves. My issue is understanding how positive charge is the same things as negative charge flowing in the opposite direction in the following situation:
Suppose there is a light bulb in the middle of the two resistors that needs a voltage drop of 20V with the battery providing 120V. If the charge flows from positive to negative, then the left resistor would need to provide the voltage drop. If the charge flows from negative to positive, then the right resistor would need to provide the voltage drop and in this case it matters which way the current is moving. I'm sure it could work either way, but my mind cannot accept that until I solve this. Could anyone help clear this up for me? 


#2
Feb1412, 07:01 PM

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P: 4,016

The current from the battery depends on the total resistance, so all the resistors contribute to the final voltage on the lamp.
So, suppose you had a lamp that needed 20 volts at 1 amp and the supply was 120 volts. The battery would still have to supply 1 amp and the resistors would have to drop 100 volts. So, the total resistance (apart from the lamp) would be 100 volts divided by 1 amp, or 100 ohms. It doesn't matter where around the circuit you put this 100 ohms and it could even be made up of a number of smaller resistors in series. You could have two 50 ohm resistors in the positions shown as R1 and R2 in the diagram. Don't get hung up about the direction of current flow. It isn't important. Just accept that current flows from positive to negative. On rare occasions you need to remember that electrons are actually flowing the other way. 


#3
Feb1412, 07:12 PM

P: 35




#4
Feb1512, 05:49 PM

P: 35

Resistor Voltage Drop Concept Question



#5
Feb1512, 06:17 PM

Sci Advisor
P: 4,016

Each resistor or lamp has a voltage across it that depends on its resistance and the current flowing in it.
So, if you had two 50 ohm resistors, each with 1 amp flowing in them, and a 20 ohm lamp with 1 amp flowing in it, then the resistors would have 50 volts across them, wherever they were, and the lamp would have 20 volts across it, wherever it was. So, the voltage drop would move if you moved the components, but the lamp would still get 20 volts to operate. It is only the potential difference across the battery that matters. Try to forget the Physics concept of "high" and "low" potential. The two terminals of a battery are equally important and neither has an inherently higher potential than the other. 


#6
Feb1512, 07:23 PM

P: 35

I do have a question about something I thought of related to Kirchhoff's Voltage Rule. In our situation, we had 120V going in and there was a voltage drop of 120V with the 2 resistors and the light. Now if the voltage drop were larger than the voltage the battery provides, would there would be no flow of current? On the other end of the spectrum, what happens when the battery provides more voltage than the circuit needs. Or can this even happen? I assume it would violate Kirchhoff's rule in a closed circuit, but what about other types of circuits? 


#7
Feb1512, 08:12 PM

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P: 4,016

No, the voltage across the lamp and resistors has to add up to the supply voltage.
The current in the circuit just depends on the voltage and the resistance, so the voltages across the resistor and lamp will vary depending on the current through them. 


#8
Feb1512, 08:16 PM

P: 35

Thank you so much for all your help. I really appreciate the time you took to explain this to me.



#9
Feb1912, 12:15 PM

P: 8

I would assume also that if you had two 50 ohm resistors and a bulb 'supposed' to run at 20 oms that if only say 110 volts were supplied then the bulb would only receive 10 volts instead of 20. Likewise if only 100 volts were supplied it wouldn't light at all.
What i can't really wrap my head around is how can a resistor placed in a circuit after a load affect the load (assuming DC at least). If you had a light bulb in a 120V circuit, I assume it would have 120V across it so that by the end of the circuit, there were 0V. If i added a second bulb in series, they would both (assuming likeness of bulbs?) then have 60V across them, so that by the end of the circuit the potential was 0. But how does that first bulb 'know' that there is a second bulb to come after it. Why doesn't it simply use up all 120V instead of 'saving' 60v for the second (forgive my inaccurate analogous terminology). 


#10
Feb1912, 01:03 PM

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#11
Feb1912, 01:55 PM

P: 628

Simon, I think therein lies the problem with your conceptualisation of electricity. An electrical component that has resistance and is carrying a current will have a voltage drop across it. It is the current flowing through the component and its resistance that defines the voltage drop. All components in a series connection experience the same current through them. You do not need to know anything more than just the resistance and the current to define the voltage drop. e.g. a 100 Ohm resistor with 1 A running through it will have a voltage drop, V=IR, = 100 V. Now you might ask what determined the current flowing around a circuit. That's easy, it is determined by the source voltage and the total (summed) resistance., V=IR again, so let's say each of your resistors are 100 Ohms, the bulb is 200 Ohm, and the source is 100 V, so the current, I=V/R, = 100 V/400 Ohm = 0.25 A. You can now inspect each of the resistances with just the 0.25 A current flowing to determine the voltage drop across each. Take the resistors, 100 Ohm x 0.25 A = 25 V. So you see that the source voltage determines the current, and the sum total of voltage drops across each resistive load must equal the source voltage, by the very definition of the voltage and the current drawn by the load. I=V[source]/[resistance 1 + resistance 2 + etc..] voltage drop across resistance X = {I x resistance X} Sum total of voltage drops = {I x resistance 1} + {I x resistance 2} + {I x etc..} = V[source] (All this relates to ohmic resistances, not reactive components, like capacitors and inductors, with AC currents .) 


#12
Feb1912, 04:29 PM

P: 35

So voltage is the amount of energy packed into 1 coulomb of charge therefore at a given component the energy it contains per 1 coulomb can vary from another component because it has to keep the same current flow, but if resistance is high and it's harder for charge to flow, more energy is needed. An analogy that helped me think of why current must be the same is thinking of people in line. It doesn't matter how fast a single person moves, they have a person in front of them and a person behind them so they all move in unison and therefore at the same speed. In a series this same logic should apply to the charges moving through it. Thank you for helping me think this through and understand it better. 


#13
Feb1912, 04:58 PM

P: 8

I think that there are 0volts at the end of the circuit is tripping me up. Or perhaps in my understanding of current and volts:



#14
Feb1912, 05:19 PM

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PF Gold
P: 11,889

You need to remember that the DC conditions refer to the steady state  after everything has settled down. At switchon, an EM pulse travels along / through the circuit and establishes the conditions that will obtain in the long term. The various Potential Differences around the circuit eventually establish themselves. Nothing 'needs to know' what it should do in advance.



#15
Feb1912, 06:20 PM

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P: 4,016

(I notice that there are two people asking here.)
I would assume also that if you had two 50 ohm resistors and a bulb 'supposed' to run at 20 oms that if only say 110 volts were supplied then the bulb would only receive 10 volts instead of 20. Likewise if only 100 volts were supplied it wouldn't light at all. Anyway, to settle this one... The total resistance is 120 ohms so if you drop the voltage to 110 volts the current will drop to 0.9167 amps. { 110 volts divided by 120 ohms is 0.9166666 amps } So the voltage across the lamp will be (I * R) or (0.9167 * 20) or 18.333 volts where it was 20 volts before. And the voltage across each of the resistors will be 45.833333 volts. If you add up 18.3333333 + 45.833333 + 45.833333 you get 110 volts. The power supply doesn't "know" about the individual resistors any more than it would feel guilty about electrocuting you. The total resistance is what determines the total current. 


#16
Feb1912, 06:30 PM

P: 35

The reason for the 0V at the very end is due to the battery, the wire actually has a voltage just like any other component as a result of the current going through it and the resistance of the wire. The battery is like an elevator that takes charges and raises them to a state of higher potential (say 12V if it was a 12V battery). You need this difference in potential to keep the current flowing. I'm not quite sure if you can say it goes from 0V up to 12V. It goes from whatever voltage it was in the wire touching the battery and is boosted up to 12V if it was a 12V battery. This 12V is only in the battery however and the voltage changes in the wire leaving the battery. The voltage in the wire leaving the battery is determined by the current that goes through it and its resistance. Remember that the current the battery provides is related to the total resistance of the entire circuit. If you have a lot of high resistance components, then the current will be slower. Less total resistance means faster current. Then again, I'm new at physics so don't take my thinking as the absolute truth until others confirm it. 


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