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I can't understand how to calculate the voltage between a and b in this open circuit. Resistances like the 9 Ohms and 13 Ohms can be ignored since no current is passing through (In case we connect a Voltmeter, we consider infinite resistance). But, how can I analise or simplify the other circuits?

enter image description here

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Hints

  1. What does the 13 ohm resistor bring to the party?
  2. What does the 9 ohm resistor bring to the party?
  3. What effect does the 18 ohm resistor have on the 15V source?
  4. What does the 4 ohm and 11 ohm bring to the party?
  5. The voltage across the 10 ohm resistor is ONLY due to which current source?
  6. Ditto the 5 ohm resistor
  7. Simplify based on these hints.

The answer is 35V from a brief mental add-up of things - it is THAT simple (assuming I've read your scrawl correctly).

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  • \$\begingroup\$ 1. Nothing 2. Nothing 3. Apparently, the 18 ohm resistor has no effect, the voltage will be 15V. But I can't understand why. 4. Same as the 3rd. The solution is correct, but I'm trying to understand part-by-part. \$\endgroup\$ – Pedro Barros Jan 19 '16 at 18:21
  • \$\begingroup\$ Nothing = because there is no current flowing through them because a and b are open circuit. No-effect because a voltage source will produce the same voltage irrespective of the load across it. \$\endgroup\$ – Andy aka Jan 19 '16 at 18:23
  • \$\begingroup\$ Your nearly there and it's right to break it down into parts. Start by redrawing with what you've learnt. \$\endgroup\$ – Andy aka Jan 19 '16 at 18:25
  • \$\begingroup\$ I have a new drawing, but I'm missing the effect 4 and 11 Ohm resistance. \$\endgroup\$ – Pedro Barros Jan 19 '16 at 18:36
  • \$\begingroup\$ Well you don't need the 4 and 11 ohm registers for calculating the potential difference between points a and b. You will get the answer even without calculating the voltage across those two registers. \$\endgroup\$ – Rahul Behl Jan 19 '16 at 18:46
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The current sources are tricky!

However, it should be straight-forward.

All of the current from the 6 A current source HAS to be going through the 10 ohm resistor. Be sure you understand why that is true before going further. The voltage across the 10 ohm resistor is \$60 \$ Volts.

Also, All of the current from the 8 A current source is going through the 5 ohm resistor. the voltage across the 5 ohm resistor is \$40 \$ Volts.

You already pointed out that the voltage across the 13 ohm and the 9 ohm resistor is zero. The voltage across the 18 ohm resistor is 15 volts by definition.

The voltage at point a with respect to point b is:

\$Vab = + 15 + 60 - 40 = + 35 \$ Volts

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  • 2
    \$\begingroup\$ Shouldn't the voltage across the 10 ohm be V = RI which is v = 10*6 = 60 Volts ? \$\endgroup\$ – Pedro Barros Jan 19 '16 at 19:24
  • \$\begingroup\$ I'd give up if I were you. For a start 8A and 5 ohms IS NOT 1.6 volts. Where did you learn ohms law? V = IR NOT V = I/R. Also 6A thru a 10 ohm resistor IS NOT 0.6 volts. \$\endgroup\$ – Andy aka Jan 19 '16 at 22:29
  • \$\begingroup\$ Oh I see you've fixed it up now. You ought to consider deleting it dude to save the embarrassment. \$\endgroup\$ – Andy aka Jan 19 '16 at 22:32
  • \$\begingroup\$ @PedroBarros it's just my personal opinion but I don't believe that fixing this travesty of ohms law is useful - if anyone upvotes it then the points go to the original author who (and again this is my personal opinion) should feel embarrassed enough to delete it. Let the original author fix it up. I'd roll-back your fix if I were you. \$\endgroup\$ – Andy aka Jan 19 '16 at 22:44
  • \$\begingroup\$ Thanks for fixing my calculations! I apologize for botching ohm's law! It really is a shame that I fumbled with the numbers... \$\endgroup\$ – Ryan Jensen Jan 20 '16 at 2:12

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