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Does the battery act like a resistor with a 10V voltage drop now?enter image description here

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  • \$\begingroup\$ You don't have a battery in your circuit. Are you referring to the 10V voltage source (in series with the 2R resistor), or something else? \$\endgroup\$
    – brhans
    Apr 17 at 15:58

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Not like a resistor, just like an ideal voltage source that maintains the same voltage difference across its terminals regardless of how much current flows into or out of it and in what direction.

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  • \$\begingroup\$ Does (positive) current flow into the positive terminal and out of the negative terminal? That is, electron current flows into the negative terminal and out of the positive terminal? Does it sustain a voltage drop across that? Will this circuit continue to work as such? \$\endgroup\$ Apr 17 at 14:42
  • \$\begingroup\$ By all of which I mean, there is still a voltage drop across the battery in this case, not a voltage rise? \$\endgroup\$ Apr 17 at 14:42
  • \$\begingroup\$ Voltage drop and rise are only relative terms based solely on the sign shown on the voltage source itself. All you know for sure is that the voltage at the + terminal is 10V more than the voltage at the - terminal. \$\endgroup\$
    – vir
    Apr 17 at 14:44
  • \$\begingroup\$ That is, the electrons in the current lose energy going from the side of the battery it enters and out of the side it leaves? \$\endgroup\$ Apr 17 at 14:44
  • \$\begingroup\$ In your example, yes. \$\endgroup\$
    – vir
    Apr 17 at 14:48

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