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Mar 21, 2022 at 10:10 comment added ringo Okay, I'll try to understand. Thank you for answer.
Mar 21, 2022 at 10:08 comment added Puk Not if it's an ideal battery: Kirchhoff's voltage law would be violated (the sum of voltage drops along the loop must be zero). It is possible if e.g. it's a 2 V battery with a 4 Ω resistor in series (which could be the internal resistance of the battery).
Mar 21, 2022 at 10:02 comment added ringo I will correct it. If the gray part is a battery connected with the + pole facing down, is this possible or not?
Mar 21, 2022 at 9:46 comment added Puk This case is certainly possible. For example, this will happen if the gray part is a 1 Ω resistor. I'm not sure what you mean by "no grayed out parts". If you mean there is nothing connected to your device (i.e. the device is open-circuited), then this couldn't happen because there would be no current flow.
Mar 21, 2022 at 9:38 comment added ringo Thank you for answer. Then, since this case is a part of the whole circuit (due to the grayed out part in the figure), is my question a possible case? Conversely, if there are no grayed out parts, is this something that can never happen? Thank you.
Mar 21, 2022 at 9:37 history edited Puk CC BY-SA 4.0
edited body
Mar 21, 2022 at 9:21 history answered Puk CC BY-SA 4.0