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Finding the current across a resistor

  1. Jan 30, 2007 #1
    1. The problem statement, all variables and given/known data
    The currents are flowing in the direction indicated by the arrows. A negative current denotes flow opposite to the direction of the arrow. Asume the batteries have zero internal resistance. Find (part 1)the current through the 5.3 ohm resistor and the 3.1 V battery at the top of the circuit and (part 2) the current through the 23.1 ohm resistor in the center of the circuit.

    Here is the circuit in question: [​IMG]

    2. Relevant equations
    V = IR

    3. The attempt at a solution
    I solved the problem but to be honest I plugged in numbers until I got the right answer. I do not understand the concept (but I would like to!) behind the problem and would really like some help decoding the problem. Here was my thought process...

    Part 1 - the overall voltage in the circuit is the sum of 3.1 V and 18.6 V. The current is flowing through from the positive termial of the 18.6 V battery, splits at the junction, and flows across the 5.3 ohm resistor. Since the battery has no internal resistance, current will be reduced across the 5.3 ohm resistor but would not be affected as it goes through the battery. Similarly, the current moves from the second battery, across the 23.1 ohm resistor and across the 5.3 ohm resistor. So, I = (18.6 V + 3.1 V)/5.3 ohm = 4.09 A

    Part 2 - The current flows from the 18.6 V battery, across the 23.1 ohm resistor. I = 18.6 V/23.1ohm = -.805 A

    I'm probably completely wrong here...but I'd really like to learn the concept behind the problem. If anyone could please explain to me where my thought process is flawed, and help me derive/understand what's going on in the diagram, I would greatly appreciate it. Thanks!
  2. jcsd
  3. Jan 30, 2007 #2


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    Homework Helper

    Problems like these are usually solved with Kirchoff's rules.
  4. Jan 30, 2007 #3
    Thats what I figured...but we havent discussed them in class so I dont really know what they are. My professor assured us in class that the problem can be solved with ohms law...but since I couldnt figure out a way to apply the ohms law, I read the book and I think I understand the kirchoff's rules, but do not know how to apply them. Junction rule states that the sum of the currents going out of a junction must equal the current going into a junction (charge is conserved) whereas the loop rule basically sates that the energy is conserved in a loop (sum of all potential differences is 0). How can I apply them here?
  5. Jan 30, 2007 #4


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    For the currents you would say that

    [tex]I_5 = I_{18} + I_{23}[/tex]

    For the potentials you would "add them up" in any closed loop in the circuit (your circuit has three closed loops). For instance the top loop would give

    [tex] - 23.1 * I_{23} - 5.3 * I_5 + 3.1 = 0 [/tex]
    Last edited: Jan 30, 2007
  6. Jan 30, 2007 #5
    I see...I guess where I'm lost is trying to figure out which way the charge is going to flow.
  7. Jan 30, 2007 #6


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    Staff: Mentor

    First of all, you should subscript the currents in the different parts of the circuit, so that you can keep track of the different values. Second, current flows through a circuit element in response to the voltage that is connected across it.

    So label the currents with subscripts and write an equation for each leg --- what do you get?
  8. Jan 31, 2007 #7


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    You are free to choose the direction of the current as you see fit. After solving for the currents you will then get either positive or negative amperes for their values. The current will flow in the opposite direction you chose it in if its value comes out negative.
  9. Jan 31, 2007 #8
    @ berkeman, So heres the updated drawing....


    If I solve for current in each branch, I get:

    I_ab = V_ab/R (but since the battery does not have any internal resistance, what would this current be?)

    I_cd = V_cd/R

    I_fe = V_fe/R

    I guess I can find the overall current in the circuit by adding up the voltage from the two batteries and the resistance from the parallel ciruits to get: I_total = V_total/R_parallel...is this useful?

    @andrevdh, that makes sense! I recall reading that in the book but it didnt click until you mentioned it.

    Now that I have an expression for currents going through each branch, how do I determine the appropriate voltage and resistance for each branch?
  10. Jan 31, 2007 #9


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    My approach (not using Kirchoff's rules) would be to notice that we have three parallel branches. The voltage over all three need to be the same, that is 18.6 V. This enables you to calculate I_cd. For the ef leg notice that the 3.1 V opposes the 18.6 V. This means that the drop over the 5.3 ohm resistor need to be 18.6 + 3.1 volt , which gives you I_fe.
  11. Jan 31, 2007 #10
    I think the prof is right, no need to resort to Kirchoffs law since there is no resistor on the lower branch of the circuit.

    the voltages across all three parallel branches must be the same and equal to the voltage of the bottom battery=18.6

    For the middle branch thats simply IR,

    For the top, its IR-3.6=18.6
  12. Jan 31, 2007 #11
    I think I understand! I get that the voltage is conserved...but why is the voltage 18.6 across all branches? Why doesnt the top battery contribute to the overall charge?
  13. Jan 31, 2007 #12
    Well they don't really but will seek to become so, in other words if you had switches involved and could instantly tie all the limbs together, they would not be at the same potential, the middle would be at zero, while the top would be at -3.6 V. The fact that there is a potential difference will cause current to flow until they are at the same potential. Convinced yet? The "bigger" reason has to do with forces on charged particles in an electric field.
    In this case the battery on the bototm is tied to the nodes with no intervening resistors so those nodes are the same potential as the battery itself. A resistor would change everything as the potential drop across it would would be unknown and would be more complicated problem.
    Last edited: Jan 31, 2007
  14. Feb 1, 2007 #13


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    Wires do not have resistance so there are no potential drop over them. This means that the three legs need to be at the same potential, which from the circuit is obvious that it needs to be 18.6 V due to the cell at the bottom. In the top leg the additional cell creates an additional potential in that leg thereby creating more potential over the resistor. The larger cell at the bottom will control the current flow in the top leg since its potential is larger, so your choice of current direction is correct. We can see that the cell in the top is assisting the current flow, thereby causing a larger potential over the resistor in that leg.
  15. Feb 3, 2007 #14
    I get it...thanks guys!!
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