In the circuit shown here the conventional current direction is from V2 to V1 and is equal to 3 Amp. I understand the theory. But, if I make this circuit with two batteries, according to the schematic the current(positive charge) leaves a positive side of one battery and goes through the positive side of the second battery.

So my question is, if I actually make this circuit with two batteries, does it mean that a battery (V1 here) accepts current to its positive side and send current from its negative side? How is that possible? How would the current circulates?


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Small, obvious, but important note for posterity (you never know who comes by in a couple of months); Apart from the notes made below by Spehro in the case of rechargeable batteries, if these batteries are not rechargeable it'll go very wrong almost from the start. \$\endgroup\$ – Asmyldof Apr 21 '16 at 17:08

You can think of this as the source V2 charging the battery represented by V1. It is absorbing energy, and V2 is providing it. The current circulates just as you said, into the + terminal of V1 and out the - terminal.

If you do an energy balance sheet, the resistor will dissipate I^2*R = 90W but V2 is providing 3A * 150V = 450W. The remainder (360W) is going into the V1 source. If V1 is a rechargable battery it may be able to absorb that for a while, but at some point it will become fully charged and the battery will get very hot and may be damaged from overcharging.


Current direction would be from right to left because sources are connected in series opposing configuration, both the current will flow in the circuit in opposite direction and produce net effect to right because of lager magnitude of right source, and infact, current would flow from 120V source, completing the path for flow but that current is not produced by 120V source.


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