Finding Current in a series circuit

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In a series circuit with a 2.0-ohm resistor and a 5.0V battery with 0.5-ohm internal resistance, the initial calculation for current (I) was incorrectly determined to be 2 amps. The presence of an ideal voltmeter, which allows minimal current flow, complicates the circuit analysis. This raises concerns about the validity of the question, as the voltmeter's function contradicts the assumption of significant current flow. The discussion highlights the need for clarity in the problem statement and suggests revisiting the question for accuracy. Overall, the circuit's configuration and the role of the voltmeter create confusion regarding current measurement.
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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.
 
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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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