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Homework Help: 2 resistors connected in parallel. same two in series what is the current?

  1. Oct 10, 2007 #1
    When two resistors are connected in parallel across a battery of unknown voltage, one resistor carries a current of 3.2 amps while the second carries a current of 1.8 amps. What current will be supplied by the same battery if these two resistors are connected to it in series?


    well the voltage received by each is the same in the parallel setup.. i thought i could do a
    IR=IR setup.. but that just confused me.. but i think i need to do something with the ratio of the currents hrmm. 1.8/3.2. but I am not sure where to go. Any hints?
  2. jcsd
  3. Oct 10, 2007 #2
    in a parallel circuit current is split up among the components.
    Voltage is shared equally by all components and equals the voltage of the source.

    so the sum of the currents for all components is the current that the source puts out.

    In series what can you say?
  4. Oct 10, 2007 #3
    in series the current is the same for each resistor.
  5. Oct 10, 2007 #4
    right, so all you have to do is to find the original current for the parallel circuit, which will be the same for each resistor when they are in series.

    You need to find the sum of the currents while the resistors are in parallel.
  6. Oct 10, 2007 #5
    well the equivalent Resistance is = 1/(1/R1+1/R2)
    but what is the equivalent current in a parallel circuit?
  7. Oct 10, 2007 #6
    the sum of the currents,

    I is split up according to the resistances. But since you already know the invidual I's for each resistor you can just add them up and get the current that the battery is putting out.
  8. Oct 10, 2007 #7
    so in a parallel circuit the to find the equivalent current you can just add them together? good to know.. 1.8+3.2= 5 amps. and they will both receive 5 amps in series? hmm
  9. Oct 10, 2007 #8
    yes, and the voltage each resistor has in series will be split up accordingly by their resistance.

    So it sort of switches from parallel/series. Voltage is the same across all components but current is split up, or I is the same and V is split up.
  10. Oct 10, 2007 #9
    Thank You :approve:
  11. Oct 10, 2007 #10
    The current will not be the same my friend.

    Think about it. Equivalent parallel resistance is smaller than the smallest resistor. which is not equal to the two resistors in series. The voltage source does not change.

    As for the original poster. The ratio of the currents will tell you which resistor is how much bigger than the other.

    if the resistors were equal, equal current will go through each. If one was twice as big as the other, the one that is twice less gets TWICE the current.

    Current always takes the path of least resistance.

    So 3.2/1.8 = 1.7778R

    That means one resistor is 1.7778 times the other (R)

    So res. 1 =1.7778R
    res 2 = R

    How do you know which resistor has which current? You tell me.
    Can you figure out the voltage across each resistor? See if you can do it.
    Last edited: Oct 10, 2007
  12. Oct 10, 2007 #11


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    There is some serious issues with the help given in this thread. bob, please stop!
    This is good as long as we remember that the total current is for the equivalent resistance of the parallel circuit. This is different from the equivalent resistance of the series circuit, so the current in the series circuit will also be different.

    Wrong! The current through each resistor will be determined by the voltage drop across that resistor in the series circuit. Each will carry the same current, but have different voltage drops. The total voltage drop will equal the battery voltage.

    Right as long as you stick with the parallel circuit, this current will not be the same in the series circuit.

    You are right to be skeptical, this is not correct.

    Ok, now let us look at a correct approach.

    we have the current in each resistor so we can write the resistance as:

    [tex] R_i = \frac V {I_i} [/tex]
    In the series circuit we have:

    [tex] I_s = \frac {V_i} {R_i} [/tex]

    We also have:

    [tex] I_s = \frac V {R_1 + R_2} [/tex]


    [tex] V = V_1 + V_2 [/tex]

    There is more then enough information in those relationships to find the series current.
  13. Oct 10, 2007 #12
    since current follows the path of least resistance, the 3.2A current goes with the R and then the 1.8A current goes with the 1.778R.
    but i cant substitute one V in for the other..? even though i know that they receive the same voltage.. or substitute R in for the other R.. when i do i get V=V which is true but it doesnt tell me what V is.. ugghh.. my head is so confused right now.. i'm sorry
    Last edited: Oct 10, 2007
  14. Oct 10, 2007 #13
    It is all relative.

    1/4 is the same as 4/16.

    So if resistor one is 1.7778R and resistor two is R, you can see that 'R' is a CONSTANT multiplier because it follows a RATIO. If 'R' is a constant multiplier then why not just assume R = 1, therefore your voltage source is 3.2 volts. If R =2, then voltage source is 6.4 volts. Doesn't matter, because it is following a RATIO. You will get the same value for the series current.

    You have the voltage source, you have the two resistor values. So 3.2V/(1.7778 ohms+ 1 ohm) = some number which will be your series current.

    do you understand? no matter what value you pick for R, everything will follow the ratio set and still give the same answer.
    Last edited: Oct 10, 2007
  15. Oct 10, 2007 #14
    ah i do see. no matter what the resistance is since they all will follow the same ratio my current in the series will be the same. just like reducing a fraction. thank you that was a much better way to explain it. if you could i also have another post which is driving me a little crazy. btw. current in the series = 1.152 amps correct?
  16. Oct 10, 2007 #15
    Exactly, it is just like reducing a fraction.
    Yes. 1.152 A is what I get as well.
    Where is your other post?
  17. Oct 10, 2007 #16
    in the same section under the heading of finding current and potential difference in a circuit
  18. Oct 11, 2007 #17


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    Did you read this?

    Start from:
    [tex] I_s = \frac V {R_1 + R_2} [/tex]


    [tex] R_i = \frac V {I_i} [/tex]
    [tex] R_1 = \frac V {I_1} [/tex]

    [tex] R_2 = \frac V {I_2} [/tex]

    [tex] I_s = \frac V {\frac V {I_1} + \frac V {I_2} } = \frac 1 {\frac 1 {I_1} +\frac 1 {I_2}} = \frac {I_1 I_2} {I_1 + I_2} [/tex]
    Last edited: Oct 11, 2007
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