Series Parallel Circuit Calculations

00:14:53
https://www.youtube.com/watch?v=9RTCYwFiHOs

Summary

TLDRLa vidéo illustre le processus de calcul de la résistance totale (RT) dans un circuit électrique complexe comportant des résistances en série et en parallèle. Elle explique d'abord comment déterminer la résistance équivalente des résistances en parallèle en utilisant la formule appropriée, puis comment combiner cette résistance avec d'autres résistances en série. Le calcul des courants dans le circuit suivant la loi d'Ohm est également abordé. Les exemples fournis montrent comment la compréhension des connexions en série et en parallèle peut aider à déterminer le comportement du circuit, notamment comment le courant se divise dans les branches.

Takeaways

  • 🔍 Étape 1 : Calculer la résistance en parallèle avant d'ajouter la série.
  • 📏 Étape 2 : Appliquer la loi d'Ohm pour déterminer les courants.
  • 📊 Étape 3 : S'assurer que le courant se divise proportionnellement aux résistances en parallèle.
  • 💡 Les résistances en parallèle rendent la résistance totale plus faible.
  • ⚖️ En série, les résistances s'additionnent directement.
  • 🔄 Le courant total est constant à travers les résistances en série.
  • 📈 La tension est partagée entre les résistances en série.
  • ⏳ Utiliser X^-1 sur la calculatrice pour les calculs en parallèle.
  • ⚠️ Faire attention à la loi de Kirchhoff pour les courants.
  • 🤓 Attention à ne pas confondre la tension totale et la tension à travers une résistance spécifique.

Timeline

  • 00:00:00 - 00:05:00

    Dans cette section, l'objectif est de calculer la résistance totale (RT) d'un circuit comprenant des résistances en parallèle et en série. La résistance de la partie parallèle (R4) est calculée à l'aide de la formule R4 = 1/(1/2 + 1/3), ce qui donne une valeur de 25 ohms. Ensuite, en ajoutant R1 qui est également de 25 ohms, la résistance totale (RT) du circuit est déterminée à 50 ohms. Ce processus simplifie le circuit pour le rendre plus facile à analyser tout en permettant de calculer le courant total (IT), qui est de 2 ampères.

  • 00:05:00 - 00:14:53

    La dernière partie se concentre sur le calcul des courants individuels (I1 et I2) à travers chaque résistance en utilisant la loi d'Ohm. En identifiant la tension et les résistances appropriées, le courant passant par R2 (I1) est calculé à 2,35 ampères et par R3 (I2) à 1,78 ampères. La somme des courants dans le circuit est vérifiée, respectant la loi de Kirchhoff, ce qui confirme que tous les calculs et analyses sont corrects.

Mind Map

Video Q&A

  • Comment calculer la résistance totale dans un circuit combiné ?

    Il faut d'abord déterminer la résistance de chaque partie (parallèle et série) séparément, puis les combiner.

  • Quel est le rôle de la loi d'Ohm dans ce calcul ?

    La loi d'Ohm permet de relier la tension, le courant et la résistance, facilitant le calcul des courants dans le circuit.

  • Comment le courant se divise-t-il dans un circuit en parallèle ?

    Le courant se divise en fonction des valeurs de résistance, chaque branche recevant une portion du courant total.

  • Pourquoi la résistance en parallèle est-elle calculée différemment ?

    En parallèle, la résistance totale est toujours inférieure à la plus petite des résistances, puisqu'on utilise la formule inverse.

  • Comment trouver le courant dans une résistance spécifique ?

    On doit calculer la tension à travers cette résistance, puis appliquer la loi d'Ohm.

  • Quelles sont les unités utilisées pour la résistance ?

    La résistance est mesurée en ohms (Ω).

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Subtitles
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  • 00:00:04
    so I hope you can see this one
  • 00:00:06
    series-parallel
  • 00:00:08
    you want to calculate R T given R 1 R 2
  • 00:00:13
    R 3 it's been parallel part series part
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    and here are our values here first step
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    to work out just the parent apart ignore
  • 00:00:24
    that part and just work out these two
  • 00:00:27
    so calculate parallel resistance total
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    of this bit here okay we'll call Rt
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    actually because it's part series
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    parallel R 4 becomes 1 over 2 plus 1
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    over 3 we put figures in
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    like so these we can do when I cannot
  • 00:01:05
    glare can be done a lot easy I know
  • 00:01:08
    because the figures I've chosen the
  • 00:01:12
    button we press on our calculator is the
  • 00:01:15
    X to the minus one so what you want to
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    do is on your calculator 1350 press X to
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    the minus one button you press plus 50 X
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    the minus one equals and then X to the
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    minus 1 again to give you the correct
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    answer and if you do all of that it will
  • 00:01:35
    come to some of you probably hope you
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    already noticed should come to 25 hours
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    this happens to me guys because I've
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    chosen it that when you have the same in
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    parallel you just take one of them and
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    then find PI the number up so there's
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    over 50 there are two of them so 50
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    divided by 2 is 25 again if we do the X
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    to the minus 1 so that's 50 X minus 1
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    plus 50x on equals x the nice one your
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    answer that will give you 25 ohms that
  • 00:02:09
    then breaks the circuit down so what
  • 00:02:13
    we've now got is this r4 which we now
  • 00:02:19
    know is 25 and r1 which was 25 as well
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    so we've now broken this circuit down
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    into just a basic series circuit and we
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    can now find our T so our T becomes oh
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    one plus r4 and r4 is this combination
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    here if we put these figures in so RT
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    for the entire circuit it's 50 ohms and
  • 00:02:58
    that can again can be drawn simplified
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    into that entire circuit
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    as so we've started off as a series
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    parallel come to a series down to one
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    these are our three steps we've taken to
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    get to our final answer Artie okay now
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    we've made the circuit into three broken
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    down pieces effectively we know Artie is
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    50
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    oh we now must do is calculate these
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    three currents this possibly makes it
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    easier to work these out this because
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    this is the total of the circuit from
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    previous the examples I T is equal to vs
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    divided by RT which is Ohm's law we put
  • 00:03:57
    these figures in 100 because this is the
  • 00:04:01
    total divided by RT just 50 is not
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    difficult to understand
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    two amps this two amps it would be the
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    same two amps that flows around in this
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    circuit here for a poorly it's this two
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    amps over here these two empty flows out
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    must come back so that's gonna to
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    amplify pass through the arm well I
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    guess at this point here it will split I
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    can split into i1 and i2 depending on
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    the value of these resistors will
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    determine which current how much current
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    goes each direction so how do we work
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    that one out
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    well to be honest there's not much to
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    work out this one because we know these
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    two are equal they're both fifty so
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    they're equal the current has to split
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    evenly therefore it will divide by two
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    so I already know and hopefully it will
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    recognize that every one and pass
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    through each one but if we need to work
  • 00:05:09
    it out we'd have to do a calculation so
  • 00:05:13
    we'd have to work out all right one
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    first hold a phone we use for that his
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    own law PV the
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    I did by ah the difference being now
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    though we want to calculate blackcurrant
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    so this step formed in this Holmes law
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    but it's wheat voltage do we apply
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    because what it isn't it isn't vs
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    because we yes is not across that
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    resistor and that's what we're trying to
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    find the current flowing through that
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    resistor there so in order to work that
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    one out
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    we've got to find the volt drop across
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    this resistor and that's easier to work
  • 00:05:52
    out from over here so to work this bolt
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    it help let's call it the one just to
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    really confuse matters to work out that
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    voltage there leave one trying to work
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    out it is Ohm's law r again I times ah
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    well looking at this circuit to find
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    that voltage there because this one is
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    all of that okay and we know from a
  • 00:06:25
    parallel circuit that voltage in a
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    parallel circuit will stay the same so
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    to work this one out it's the current
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    passes through that so this one here
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    it's two amps because it's a combination
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    of those two so becomes my teeth over
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    here times are four okay we know he's 25
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    so that becomes 2 times 25 50 volts okay
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    numbers now to be simple so that becomes
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    one that's one will be an only divided
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    by R well there's actually one two three
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    four five different hours on here
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    because we want to find the current
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    through that one its car to so that
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    becomes r2 which is 50 55 50
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    is one amp that's occurring there this
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    one can be done exactly the same way if
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    you needed to calculate it because it
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    would be I 2 is equal to V 1 this time I
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    divided by R 3 so they wanted curd it's
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    the same thing becomes 50 again and it's
  • 00:07:54
    divided by 50 and that is also why back
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    then means is 2 amps will come up it
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    will split evenly 1 amp goes without 1 1
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    it gets one and there's your answers to
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    those things there okay let's try it
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    again with a different set of values
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    Maurice Nicoll suppose in a sense it's
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    harder to work out 10 25 33 so as before
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    we will need to calculate the parallel
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    part first and remember two plus one
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    over r3 put calculators ultimately 33 as
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    well this way we can use that calculator
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    and remember the bottom we use the exit
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    - 125 X -1 + 33 X the minus 1 equals
  • 00:09:13
    okay set fractions most calculators need
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    to do that now we must to X the mods 1
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    your answer excellent 1 again press the
  • 00:09:22
    equals come down a fraction yesterday
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    comes out there 14.2 okay
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    you
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    our original one was 10 so we can then
  • 00:09:54
    work out Artie Artie becomes R 1 plus R
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    4 because we're now looking at the
  • 00:10:04
    series part and if we put these figures
  • 00:10:07
    in 10 plus
  • 00:10:18
    don't read the unit so that is now Artie
  • 00:10:33
    how can we now do we can now work out
  • 00:10:35
    the currents for these ones here
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    okay so work out the currents then and
  • 00:10:42
    work out my t i1 and i2 as before so
  • 00:10:47
    let's start with my tea
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    that's nice simple one because we can
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    just look at this here Emma IT is here
  • 00:10:59
    here because these are the same circuits
  • 00:11:03
    I've been working out here therefore
  • 00:11:06
    mighty fusing vs divided by RT so total
  • 00:11:14
    total and a total 20 2,400 divided by 2
  • 00:11:32
    and if we did it into the calculator we
  • 00:11:34
    will get 4 point 1 3 amps so that then
  • 00:11:43
    is the current flowing through in all
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    these locations so mighty four point 1 3
  • 00:11:52
    amps we can now work out I 2 and I 1
  • 00:12:01
    they are done exactly the same way as we
  • 00:12:04
    did before so let's do I 1 which will be
  • 00:12:08
    Ohm's law they divided by R but it's
  • 00:12:12
    which V so it's not vs remember because
  • 00:12:14
    it's not a total which one I work out it
  • 00:12:16
    has to be the voltage that is in this
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    section of the circuit so we'll call
  • 00:12:21
    that 1 V 1 and we want to find that
  • 00:12:29
    current there so he's r2 so we've got to
  • 00:12:34
    work out V 1 which the moment we don't
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    know r2 we do know what that is that is
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    25 so we've got to find V 1 so let's
  • 00:12:43
    find V 1 V 1
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    is I times R from straightforward law
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    but is which current or we only know
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    that current can't use this because we
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    got current through that we can look at
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    this circuit because this here is all of
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    that so that becomes I T times our force
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    look at the series we made up so that
  • 00:13:09
    becomes four point one three times our
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    four G's fourteen point two two and if
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    we pull out to a character later via
  • 00:13:31
    fifty eight point seven three rounded up
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    and this will be volts so that then
  • 00:13:41
    these are V 1 so that is what goes up
  • 00:13:44
    here fifty eight point seven three we
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    then do 58.7 3/25 yet 2.35 amps in that
  • 00:14:04
    part we can now work out what I 2 will
  • 00:14:07
    be the last time we did I draw again
  • 00:14:11
    however weekend is we can apply
  • 00:14:13
    Kirchhoff's current law and that I T
  • 00:14:17
    minus I 1 so we know I t's 4.13 we take
  • 00:14:26
    away the upper current which is two
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    point three five
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    is 1.78 so there's our three currents
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    flow around in this circuit
Tags
  • résistance totale
  • circuit en série
  • circuit en parallèle
  • loi d'Ohm
  • courant électrique
  • résistances équivalentes
  • calcul de courants
  • tension
  • électrique
  • énergie