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Homework Help: Finding voltage drop in parallel and series circuits

  1. Nov 8, 2004 #1
    i think i may have the solution to this problem, but something about it is still bothering me.

    the question is, if you have a 100W and a 60W bulb, both on the same series circuit, across which bulb will there be a greater voltage drop? if the same are on a parallel circuit?..

    as i understand voltage drop, it is proprtional to resistance. so,

    Watts = amperes x volts


    watts = (volts / ohms) x volts

    then solve for resistance. but how would these be any different on a series or parallel?

    :grumpy: i'll admit right now that i am more than a little confused as to what voltage drop really is. :grumpy: please clarify!!any help would be greatly apreciated!
    Last edited: Nov 8, 2004
  2. jcsd
  3. Nov 8, 2004 #2
    Okay first off, when a circuit element such as a resistor is connected to a source of emf (a battery for instance), current flows through it. One end of the resistor connected to the +ve terminal of the battery is said to be at a higher potential with respect to the end connected to the -ve terminal. So when current flows from +ve end to the -ve end, the potential is said to drop. As a concrete example, if you have a 15 V battery connected to a 10 ohm bulb, the potential drop across it is simply 15V. However, if I connect a 20 ohm bulb in series with, the potential drops (which must add up to 15V due to conservation of energy) are 10V and 5V across the 20 ohm and 10 ohm resistors, respectively.

    Back to your problem...what happens when the bulbs are in series? What parameter is same for both of them? What about when they are in parallel. Their resistances are (obviously) fixed so what else can change?

  4. Nov 8, 2004 #3
    Disclaimer: I wrote what I wrote below mainly to see if I could construct a plausible description of voltage and what goes on in circuits for myself, so it's a bit over-the-top. Also, some of it is written to stimulate intuition and is somewhat inaccurate to the gory details of scientific truth (I'm referring to where I'm describing the movement of charge down a wire and such). Still, this should hopefully help you get at least some idea of how voltages/currents work. The inaccuracies in qualitative description do not translate to inaccuracies in equations.

    Voltage is basically a measure of how much energy a unit charge will dissipate going through the circuit. W=Vq, where W - work done by given charge, q - magnitude of the charge. If a wire was a hollow tube with nothing in it except a single electron speeding along, the electron would acquire a kinetic energy at the end of the tube that would be equal to the voltage times its charge. However, that wire in reality is chok-ful of particles and such. Individual electrons and and also the positive charges present keep butting each other and don't accelerate very far; instead they're bouncing around, hitting other charges, rebounding, hitting others again and so forth.

    Since a voltage _is_ being applied across the circuit (there is an electric field in the wire), though, this bouncing around does result in a very slow net flow of charge down the wire (the electric field accelerates the charged particles when they're bouncing in one direction and decelerates them in the other, the net effect being movement in one direction). The funny thing is, it turns out (there's a lot of math behind this) that the _rate_ of the flow of charge is the same throughout the wire/circuit (in other words, the net flow of charge is not accelerating across the circuit). The kinetic energy that electrons would acquire by accelerating freely down the imaginary "empty wire" now shows up as the random bouncing-around kinetic energy energy of the charges as they move down the wire. The greater the voltage and hence the electric field, the faster the charge flows (=greater the current I, since the current is the charge passing through any crossection of the wire in one unit of time), the faster the electrons bounce around (since the field accelerates them more) . But all that bouncing of particles is really the same thing as heat - the voltage is effectively heating up the wire. And heat is energy - so the voltage is transforming electric energy into heat energy. (That is then subsequently absorbed by the environs, which is why a heater warms you or why you see a lightbulb glow).

    So the voltage drop across a circuit measures how much energy would be dissipated by a unit charge going through that circuit. Hence the familiar formula P=IV - power is energy per time and current is charge per time, so if you multiply both sides by time you get W=QV, which says the energy dissipated (work done by charge moving) is equal to the charge that has moved through the circuit (effectively the amount of charge that has passed any point in the circuit in that time) times the voltage across the circuit. Furthermore, for an (non-existent) ideal voltage supply that can supply infinite current, that energy (and hence voltage drop) is always going to be the same, no matter what resistor you plug into the circuit. (Note that this is not the same as saying that the power is always going to be the same).

    Similarly, a voltage drop across a portion of a circuit is the energy dissipated by a unit charge going through that portion of the circuit.

    So what happens when you put some resistor in the circuit? Well, resistors are basically bottlenecks of current - they make it hard for current to pass through. This forces charge to slow down (which decreases the current). In fact, it turns out that, to a good approximation, the current is inversely related to a certain property of a resistor called R for resistance, which is a function of the physical properties and dimensions of the resistor. This is Ohm's Law: I=V/R. That is, if you put a tighter bottleneck into the circuit, current is going to be slowed down more - R has increased. (Incidentally, when there are no explicit resistors in the circuit, the wires themselves are the only resistors - the only thing that impede current. R is very low for wires, so I gets very high and so does the power P=IU. Hence the danger of short-circuiting).

    Now to your circuit. When you have two resistors strung in parallel, it's as if there were two bottlenecks for the current to get through, one after the other. Their effect is cumulative, so you can write Rtotal=R1+R2, where Rtotal is the total resistance of the current. Why? Well, think of it as a traffic jam: first some people hit the first congested spot, slow down, then hit the next one right after and slow down even more, and soon there's cars slowed down to the same snail's pace all behind them. Now if you use Ohm's law you'll see that the current drawn will be I=Vtotal/(R1+R2). However, you want to know the voltage drop across each of the particular resistors. Here again you use Ohm's law. Now you know the total current that's going to be drawn by the circuit, so you can calculate how much energy that current will dissipate going through a particular resistor (bottleneck). For instance, V1=IR1=Vtotal*R1/(R1+R2). Intuitively, this means: the narrower the bottleneck R1, the harder the charge will have to squeeze to get through, so it'll heat the resistor up more (and dissipate more energy, leading to a greater voltage drop).

    When you put the two resistors in parallel, it's a bit different. The current splits into two where the resistors branch off. One of them now has to squeeze through a tighter bottleneck than the other (higher R), so the current there will be slowed down more than in the other branch (so I will be lower). But - and this is the crucial point - the voltage drop across both branches _must be the same_ and equal to the voltage drop from the place where the current branched out to where it comes back together. Why? Well, putting it in fancy language, integrals across the electromagnetic field are path independent - that is, if you start from one point and go to another point, the energy it takes you to do so is always the same no matter what path you took. More intuitively, you could have a perpetual energy source if the voltage drop across one branch was bigger than that that across the other - you could drive current down the bigger voltage path, drive it up the smaller voltage path (push it back through the resistor), return to where you started and have a net gain in energy (you'd have gained spare energy by going in a loop). This is not all that intuitive and is worthy of a lot of attention, by the way. But anyway, the voltage drop across both branches _is_ always the same, so you can write I1=V/R1 and I2=V/R2; Itotal=V/R1+V/R2=V(1/R1+1/R2). This tells you that the two resistors in parallel branches act as one resistor for which 1/Rtotal=1/R1+1/R2. But you don't need that for this problem; you only need to note that the voltage drop across two branches in parallel is always the same and you're done (the first part of the problem was discussed above).

    Hope this helped some.
  5. Nov 8, 2004 #4
    let me try and summarize, see if i get this or not.

    series circuit with a 9 volt sorce and one bulb on it has a 9 volt drop, both on the circuit as a whole and across the bulb.

    parallel circuit, 9 volt, with 2 branches has a (1/2 * 9 volt) drop across each branch. 3 branches would be (1/3 * 9V) @ each branch.

    series circuit with a 1 ohm resistor and a 2 ohm resistor would be 3V drop and 6V drop respectively.

    but assuming that i don't have the ohms of the resistors, and instead have the watts, how do i get there from here? do i:

    Watts = amperes x volts

    which means

    watts = (volts / ohms) x volts

    and isolate resistance? then set up the ratio for V drop based on resistance? i assume it really doesn't matter what the voltage is (as it wasn't given, neither was resistance, in the problem), it could be any number as it is identical for everything on the circuit and won't be chaged, right? what i mean is, the voltage across each device is the same, the resistance is specific to each device.

    oy vay! i am sorry, i have been thinking about this problem for so long now that it seems to be making less sense all the time. i understand, i think what voltage drop (or potential difference) is when we talk about the circuit as a whole. i guess i still can't vizualize what it means when we talk about it in relation to the devices on that circuit. it seems to mean a ratio of potential based on resistance.
  6. Nov 8, 2004 #5
    The voltage drop across three branches is the same _and equal to the voltage drop across the whole circuit_. You're just splitting the charge into three branches, but a unit of charge, whichever branch it goes through, must still dissipate the same energy getting from the start of the circuit to the end. It's like if you had a waterfall split into three and then come together again - each individual water droplet would've accelerated and gained the same kinetic energy if the waterfall had never split at all. So if you have three branches and a total voltage of 9V across the circuit, the voltage across each branch is 9V.

    (of course, when you say that a voltage drop across a part of a circuit is the same as the voltage drop across the whole circuit, that's always an approximation since there's always some small voltage drop in the wires, but it's a good approximation to make in most cases)
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