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Current question

  1. Dec 4, 2007 #1
    [SOLVED] Current question

    Hey, I have been working on this electric circuit section in class for a couple of days. I was just wondering if I am on the right track with this question seems how my teacher won't give me a very straight answer.

    Six volts is applied to a 2 ohm resistor in series with two 2 ohm resistors connected in parallel. The current through the first resistor is?

    1/R = 1/2 + 1/2
    1/R = 2/2
    R = 1


    R total = 1 + 2 = 3 ohms

    I am not sure if I should even find the total reistance or can I just take
    I = 6V / 2ohms
    I = 3A
    Is that correct? Or am I even close?
  2. jcsd
  3. Dec 4, 2007 #2


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    Homework Helper

    Is the total resistance 3 ohms? In which case,

    I = 2 A

    Yes, you want to reduce the series-parallel network to a single "equivalent resistance" in order to find the current through the network. So that current is 2 A.

    All of that current must go through the first resistor, so the current through it is also 2 A.

    You can now check this. Since 2 A goes through the 2 ohm resistor, the voltage drop across it is 2 A · 2 ohm = 4 V. The current splits evenly through the two parallel 2 ohm resistors, so each one passes (2 A)/2 = 1 A of current; the voltage drop across each will then be
    1 A · 2 ohm = 2 V. The 4 V drop across the first resistor plus the 2 V drop across the set of parallel resistors adds up to 6 V, which is indeed the applied voltage.
    Last edited: Dec 4, 2007
  4. Dec 4, 2007 #3
    Yes but wouldn't I=6V/2ohms give a 3Amp current? 6/2=3?
  5. Dec 4, 2007 #4


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    The single 2 ohm resistor is not the only resistance in the circuit. The energy from the voltage source that makes it possible to establish the electric field that forces the charges along the circuit must act through all of the resistances (this is a slightly glib way to put things), so the full effective resistance of all the components in the circuit (including the wires -- though we ignore this in introductory courses) must be considered.

    It will turn out that six volts is not acting across that first 2 ohm resistor; it will only be four volts. The voltages across the assorted components in the circuit varies; it is not 6 V across each device...
    Last edited: Dec 4, 2007
  6. Dec 4, 2007 #5
    Okay thank you i understand now
    Last edited: Dec 4, 2007
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