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In class I had to analyze the following circuit: enter image description here

When I solve the circuit, using superposition, I get there is a current flowing thorough the ideal voltage source in the right. What will the voltage source in the right actually do? Will it become like a resistor and cause to a voltage drop of 4v?

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    \$\begingroup\$ Ideal voltage sources are not necessarily power sources. Ideal voltage sources constrain the voltage between two points. Current may flow in either direction through them. They may provide power to the circuit or absorb power from the circuit. \$\endgroup\$
    – Elliot Alderson
    Commented Mar 26, 2019 at 12:01

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You need to define a reference point, I've put it below the 50V source. The 4V source does not actually behave like a resistor, it doesn't influence the current due to its internal resistance but rather just 'reduces' the overall applied voltage (what of course also influences the current). Due to this, the resulting current flowing from the positive of the 50V to the negative is

\$(50V-4V)/27\Omega=1.7037A\$

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ I have one question, why is important to have a reference point? \$\endgroup\$
    – Idan Aviv
    Commented Mar 26, 2019 at 8:47
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    \$\begingroup\$ @IdanAviv The ground reference doesn't change anything to (the behavior) of the circuit itself. It is just a "trick" we EEs like to use so that we can easily say: the voltage at the left side of R1 is 50 V. Without the ground reference we cannot say that, we'd have to say: The voltage at the left side of R1 and the - contact of V1 is 50 V. Note how I need to specify two points instead of only one. The ground reference is always (by definition) 0 V so it suffices to name only one node, the fact that the other node is 0 V is implied. \$\endgroup\$
    – Bimpelrekkie
    Commented Mar 26, 2019 at 8:55
  • \$\begingroup\$ Now I get it, thank you both !! \$\endgroup\$
    – Idan Aviv
    Commented Mar 26, 2019 at 8:57
  • \$\begingroup\$ @IdanAviv Maybe later in life you will simulate a circuit in a circuit simulator, then if you forget the ground symbol, you might encounter all kind of weird behavior. Having no explicit ground reference confuses many circuit simulators (yes even the expensive ones). So it is better to get used to it now and always add a ground symbol. \$\endgroup\$
    – Bimpelrekkie
    Commented Mar 26, 2019 at 8:58
  • \$\begingroup\$ I have another question, what happened if I connect the two power sources without a resistor between them (like R1=R2=0). Will the voltage source behave the same? \$\endgroup\$
    – Idan Aviv
    Commented Mar 26, 2019 at 9:03
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No, it's not a resistor. It's a creature that current flows in it independently of its own voltage. The voltage is just given, and current flows, so all the mesh behaves according to Kirchhoff's law.

Imagine a battery. A stronger battery can still push electrons towards it, that doesn't change the voltage of the battery itself.

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  • \$\begingroup\$ Current is not independent of the 4 V source. Both sources affect the current. \$\endgroup\$
    – Chu
    Commented Mar 26, 2019 at 10:20
  • \$\begingroup\$ So? So does the load... Which is according to the kirchhoff's law. \$\endgroup\$
    – user76844
    Commented Mar 26, 2019 at 10:24
  • \$\begingroup\$ Perhaps 'current flows in it independently of its own voltage' is misleading. \$\endgroup\$
    – Chu
    Commented Mar 26, 2019 at 10:30
  • \$\begingroup\$ No. 1234567890ABCDEF \$\endgroup\$
    – user76844
    Commented Mar 26, 2019 at 10:38

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