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My wire is running 5 volts, 2 amps. I am splitting that wire into two. I want to make sure 1 amp goes to each half. According to Kirchhoff's law, that is possible, because it states that the sum of currents through all wires on the receiving side of the junction (node) is equal to the sum of currents through all wires on the supplying side.

However, I don't understand it properly. What is the "loop" in Kirchhoff's voltage law? Is there a difference between KVL and KCL?

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    \$\begingroup\$ The trick is to ensure that the load on each wire is exactly 5 ohms so it takes exactly 1A at 5V. If you can't guarantee that each load is 5 ohms, you'll have to describe what you're really trying to do. \$\endgroup\$
    – user16324
    Dec 27, 2016 at 13:57
  • \$\begingroup\$ thanks so i add a 5 ohm resistor on the two halves o the wire \$\endgroup\$
    – Ali Ragb
    Dec 27, 2016 at 15:19
  • \$\begingroup\$ By the way, it's written Kirchhoff's law. You will find better information about it using the correct spelling. \$\endgroup\$
    – Janka
    Dec 27, 2016 at 16:52
  • \$\begingroup\$ @AliRagb "add a 5 ohm resistor..." - no. Why are you splitting the wire into two? \$\endgroup\$ Dec 27, 2016 at 19:39
  • \$\begingroup\$ @AndrewMorton because i have two amps 5 volts, and i want to power two 5 volt 1 amp devices \$\endgroup\$
    – Ali Ragb
    Dec 29, 2016 at 14:25

1 Answer 1

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There are two basic ways you can do this.

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By Ohm's law if you have a current of 2A from a voltage of 5V then the load must equal 2.5 Ohms.

In circuit A if we add an identical piece of wire (same material, cross-sectional area and length) between X and Y the current will split equally (2A --> 1A + 1A) between the two wires at point X. At point Y the two currents will recombine back to the 2A (1A + 1A --> 2A).

Alternatively (circuit B) you could double the load resistance to form two parallel loads of 5 Ohms. Two 5 Ohms in parallel giving the original load of 2.5 Ohms. Feed each 5 Ohm with a separate wire from point X. In which case the 2A current will split at point X (2A --> 1A + 1A) and 1A will pass through each 5 Ohm load with the currents recombining at the common load connection before returning to the negative terminal of the battery.

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  • \$\begingroup\$ @Jlm Dearden but is it going to split exactly , or is it not. Wouldnt adding two 5ohm resistors going to help let exactly 5 volts one amp? \$\endgroup\$
    – Ali Ragb
    Dec 29, 2016 at 14:30
  • \$\begingroup\$ The two (identical) 5 Ohms are in parallel making it the equivalent of 2.5 Ohms. This makes for two current loops of 1A each giving the original 2A total current. It will split exactly provided that the two resistors (circuit B) or wires (circuit A) are identical. \$\endgroup\$ Dec 29, 2016 at 14:52

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