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I need to determine the current across the 4 ohm resistor (name it I1).

What I calculated, thought out so far :

  • Current of 2 ohm resistor : I=5V/2ohm=2,5A. I Think this current is flow towards the A point (because the current and voltage direction is always the same on passive components, is it true?)
  • The currents on the A point is : 5A+2,5A=7,5 A comes in , and 3.5 flows out. 7,5A-3,5A=4A
  • This means 4A needs to flow out from the A point , i think it's flow toward the one ohm resistor.
  • Voltage on 1 ohm resistor is U=4A*1ohm=4V

I cannot calculate from there the current through the 4 ohm resistor.

  • \$\begingroup\$ When you use Ohm's law the current is assumed to enter the end of the resistor that has the higher voltage. Your diagram has arrows next to voltage values...that is non-standard and confusing. \$\endgroup\$ Commented Nov 6, 2018 at 15:38
  • \$\begingroup\$ hint: what is the voltage between node A and the top node (+ side of the 6V voltage source)? \$\endgroup\$
    – jsotola
    Commented Nov 6, 2018 at 17:10

1 Answer 1


You're almost there. Use KVL to find the voltage on either side of the \$4\Omega\$ resistor. Because you know the voltage across the \$1\Omega\$ resistor, you can solve for the voltage across the resistor you care about by applying KVL.

  • \$\begingroup\$ There is no "voltage at either side" without specifying a common reference point (ground). KVL will give the voltage across the resistor. \$\endgroup\$ Commented Nov 6, 2018 at 15:30
  • \$\begingroup\$ I tried the kirchoff's law, it gives out 0v if I sum the voltages inside the "triangle". I'm really confused about this exercise. \$\endgroup\$
    – T.Balazs
    Commented Nov 6, 2018 at 15:48
  • \$\begingroup\$ @ElliotAlderson Oops, yes, I see how that could be confusing. Will edit accordingly \$\endgroup\$
    – Shamtam
    Commented Nov 6, 2018 at 16:24
  • \$\begingroup\$ @T.Balazs Okay, and using Ohm's law, what does \$I\$ equal if \$V = 0\$? \$\endgroup\$
    – Shamtam
    Commented Nov 6, 2018 at 16:25
  • \$\begingroup\$ I equals V/R ,then 0volt/4ohm will be 0 Amper \$\endgroup\$
    – T.Balazs
    Commented Nov 6, 2018 at 17:32

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