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Essentials
Forum PARALLEL CURRENT
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Related

PARALLEL CURRENT

salesm21
salesm21 over 4 years ago

Recently been thinking about the ways current flows through a parallel circuit and it confuses me. So all connections on a parallel circuit are at equal potentials with the source. So no two components are getting anymore or less energy than the others. However based on the resistive nature of that branch will determine the current that is pushed through that part. So id like to simplify this by taking two circuits with equivalent parts. Both circuits have a DC source of 10V and each has 5 resistors equaling 5 ohms each. In series each resistor would draw away not current but the energy that is propelling electrons down the conductor. The circuit has .4A of current.  In a parallel circuit all five have 10volts dropped on them and therefore each has its own current (In this case they would all be identical but each is drawing a separate current.) My question is where does all the extra current come from? Is it front the extra paths or was it always in the source?

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  • genebren
    genebren over 4 years ago +4 suggested

    All current comes from the source that is providing the potential (assuming that it is an ideal source, with no series resistance).

     

    In the initial steps of understanding electronics the supply is ideal…

  • mcb1
    mcb1 over 4 years ago +3 suggested

    My question is where does all the extra current come from?

    Ohms law assumes the supply voltage and current is whatever your figure ends up.

     

    5 x 5ohm resistors in series across 10v will be I=E/R = 10/25 …

  • jw0752
    jw0752 over 4 years ago +1 verified

    Hi Mitchell,

    Many years ago when I was just starting in electronics I first encountered Ohms law. Using the formula I determined that I should be able to get an awesome flash of current by putting a dead…

Parents
  • mcb1
    0 mcb1 over 4 years ago

    My question is where does all the extra current come from?

    Ohms law assumes the supply voltage and current is whatever your figure ends up.

     

    5 x 5ohm resistors in series across 10v will be I=E/R = 10/25 = 0.4A as you calculated.

    There is only one path for current to flow which is throught the first resistor, then the second and so on.

    If you measured the voltage across each resistor it would be 2 volts as it gets 'spread' across each resistor.

     

    BUT 5 x 5 ohm resistors in parallel means there are 5 seperate circuits that are across the 10v source.

    The current in each path is I=E/R = 10/5 = 2A, but there are 5 paths so the total current is 2 x 5 = 10A

    The equivalent resistance is 1/Rt = 1/R1 + 1/R2, etc  = 1/5 + 1/5 + 1/5 + 1/5 + 1/5 = 1 so 1/Rt = 1 which equates to 1.

    I=E/R 10/1 = 10A.

     

    The picture and calculations here might help.

    https://www.swtc.edu/Ag_Power/electrical/lecture/parallel_circuits.htm

     

     

    So these calculations show what the expected current will be flowing in the circuit using these values.

    If it was implemented, you would require a supply capable of supplying 10v at 10 Amps.

     

    Mark

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  • mcb1
    0 mcb1 over 4 years ago

    My question is where does all the extra current come from?

    Ohms law assumes the supply voltage and current is whatever your figure ends up.

     

    5 x 5ohm resistors in series across 10v will be I=E/R = 10/25 = 0.4A as you calculated.

    There is only one path for current to flow which is throught the first resistor, then the second and so on.

    If you measured the voltage across each resistor it would be 2 volts as it gets 'spread' across each resistor.

     

    BUT 5 x 5 ohm resistors in parallel means there are 5 seperate circuits that are across the 10v source.

    The current in each path is I=E/R = 10/5 = 2A, but there are 5 paths so the total current is 2 x 5 = 10A

    The equivalent resistance is 1/Rt = 1/R1 + 1/R2, etc  = 1/5 + 1/5 + 1/5 + 1/5 + 1/5 = 1 so 1/Rt = 1 which equates to 1.

    I=E/R 10/1 = 10A.

     

    The picture and calculations here might help.

    https://www.swtc.edu/Ag_Power/electrical/lecture/parallel_circuits.htm

     

     

    So these calculations show what the expected current will be flowing in the circuit using these values.

    If it was implemented, you would require a supply capable of supplying 10v at 10 Amps.

     

    Mark

    • Cancel
    • Up +3 Down
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    • Verify Answer
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