Question: http://imgur.com/a/ArcyP

The problem is I always stuck at this kind of problems can you tell me how to solve this problem ?

I think about using Thevenin or Norton but circuit seems to confusing.

  • \$\begingroup\$ Where do you stuck? \$\endgroup\$ Jan 7, 2017 at 16:42
  • \$\begingroup\$ Please show your calculations so someone might spot a mistake. \$\endgroup\$ Jan 7, 2017 at 16:44
  • \$\begingroup\$ Simplify everything you know how to, then come back after you've shown some effort. \$\endgroup\$
    – Daniel
    Jan 7, 2017 at 16:45
  • \$\begingroup\$ If you straighten out the parts drawn at funny angles, the circuit will appear much simpler - the three resistors on the left are all in parallel. \$\endgroup\$ Jan 7, 2017 at 16:45
  • \$\begingroup\$ @PeterBennett yeah i realised that but i have questions like : 1) Can i eliminate 2 voltage sources at left like there is 2(+_-) and 1(-_+) voltage source are those eliminate eachother ? 2) if we use thevenin and make current sources open terminal would current still go through the resistors (50-100-150 the ones with on the same cable with current sources) ? \$\endgroup\$
    – onur cevik
    Jan 7, 2017 at 16:52

1 Answer 1


This is actually a very simple circuit, after making use of the following facts:

  • Elements in parallel to a voltage source can be omitted
  • Elements in series with a current source can be replaced by a short circuit

The voltage sources set the node voltage at node A to -5V, since two of them cancel each other.

At node B we have 5 currents, three are known. The other two can be written using the two node voltages VA and VB and the resulting equation can be solved. VA is already known, so only VB needs to be calculated.

The only resistors to care about are R4 and the two 50ohm resistors to the right. Everything else is unimportant.

  • \$\begingroup\$ yeah but why we omit the resistors that series with current source if there is a line like (Current source)---------Resistor-----Ground wouldnt current flow through resistor ? \$\endgroup\$
    – onur cevik
    Jan 7, 2017 at 17:14
  • \$\begingroup\$ intermediate voltage is irrelevant \$\endgroup\$ Jan 7, 2017 at 17:19
  • \$\begingroup\$ @onur cevik -- That was a bit sloppy. The series resistor can be replaced by a short circuit. \$\endgroup\$
    – Mario
    Jan 7, 2017 at 17:34
  • \$\begingroup\$ @Mario sorry my point was why are we doing that ? I mean why we are making it short circuit ? Wouldn't current from the source go through resistor. sorry for my english. \$\endgroup\$
    – onur cevik
    Jan 7, 2017 at 17:41
  • \$\begingroup\$ Look at the branch with the 20mA current source. It injects a current of 20mA into node B and it draws a current of 20mA from the ground node. The resistor is unimportant, because the current source already sets the branch current. Whatever resistors value you would pick, the current source would account for it by adjusting its voltage. \$\endgroup\$
    – Mario
    Jan 7, 2017 at 17:44

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