Electric circuits voltage flow question

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SUMMARY

The discussion centers on the behavior of voltage readings in an electric circuit with a focus on V2 and its relationship to the battery voltage of 12V. When the switch is open, there is no current flowing, allowing V2 to display the same voltage as V1, which is 12V. Conversely, when the switch is closed, the circuit becomes a short circuit, resulting in a voltage reading of 0V at V2 due to negligible resistance in the wires. The application of Ohm's Law (V=IR) confirms these observations.

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  • Understanding of Ohm's Law (V=IR)
  • Basic knowledge of electric circuits and components
  • Familiarity with voltage, current, and resistance concepts
  • Experience with circuit diagrams and analysis
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  • Learn about Kirchhoff's Voltage Law (KVL)
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TheRedDevil18
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Homework Statement



circuit.jpg


Homework Equations





The Attempt at a Solution



12.1.1) I don't understand why V2 would have a reading of 12V when V=IR and there is no current flowing in the circuit, can someone please explain?

12.1.2) I think I understand why the voltage is 0 and that is because the resistance in the wires are negligible and voltage is dependent on resistance, Correct?
 
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1) Since there's no current flowing when the switch is open, all resistors can be disregarded. When you redraw the circuit with no resistors, V2's leads will be connected to the anode and cathode of the battery. Therefore, V2 should display a reading of 12V same as V1.

2) Yes, when the switch is closed (short circuit), the reading on V2 should be zero since the wire's resistance is negligible, or in other words, zero. Applying V=IR will result in V= 0.
 
Abstr7ct said:
1) Since there's no current flowing when the switch is open, all resistors can be disregarded. When you redraw the circuit with no resistors, V2's leads will be connected to the anode and cathode of the battery. Therefore, V2 should display a reading of 12V same as V1.

2) Yes, when the switch is closed (short circuit), the reading on V2 should be zero since the wire's resistance is negligible, or in other words, zero. Applying V=IR will result in V= 0.

Ok, that makes sense, Thanks for the explanation
 

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