Finding Current in a series circuit

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SUMMARY

The discussion centers on calculating current in a series circuit involving a 2.0 Ohm resistor, a 5.0V battery, and an internal resistance of 0.5 Ohm. The initial calculation using the formula V=IR led to an incorrect conclusion of 2 amps. Participants highlighted that a voltmeter, designed to measure voltage with minimal current flow, complicates the scenario, indicating that the question lacks clarity and completeness. The consensus is that the circuit's configuration and the role of the voltmeter render the problem unsolvable with the provided information.

PREREQUISITES
  • Understanding of Ohm's Law (V=IR)
  • Knowledge of series circuit principles
  • Familiarity with voltmeter functionality
  • Basic circuit analysis skills
NEXT STEPS
  • Review the principles of series circuits and current flow
  • Study the characteristics and limitations of voltmeters
  • Explore circuit analysis techniques for complex circuits
  • Learn about the impact of internal resistance in battery circuits
USEFUL FOR

Students studying electrical engineering, educators teaching circuit theory, and anyone interested in understanding the nuances of series circuits and measurement tools.

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Homework Statement



An ideal voltmeter is connected to a 2.0 Omega resistor and a battery with emf 5.0V and internal resistance 0.5 Omega

Series Circuit

YF-25-35.jpg


Homework Equations



V=IR
V=I(R1+R2)
R(total)=R1+R2

The Attempt at a Solution



5.0V=I(.5 Ohms + 2.0 Ohms)

solve for I because current is conserved through a series circuit

I=2amps...this is wrong.
 
Last edited:
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There is something very odd about this circuit. A voltmeter is designed to only let a tiny current (ideally just one electron per second) through in order to measure the energy per charge (Voltage). Thus practically no current will flow around the circuit.

It isn't a sensible question with the information given. Better check the question again. Incidentally, you didn't list any actual questions.
 

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