What is the resistance of the voltmeter in this series circuit?

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thereddevils
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Homework Statement



A 500 ohm resistor and a 1000 ohm resistor are connected in series with a 12 V battery of negligible internal resistance . When a voltmeter is connected across the 500 ohm resistor , its reading is 3 V . What is the resistance of the voltmeter ?


Homework Equations



V =IR

The Attempt at a Solution



the effective resistance across the voltmeter and 500 ohm resistor (i am not sure across which) is Re=500Rm/(500+Rm) where Rm= resistance of meter and Re=effective resistance

From V=IR , 12=I(1500) , so the current flowing through the circuit is 0.008 A .

3=0.008(500Rm/(500+Rm))

solving this , Rm=1500 ohm

but the answer given is 1000 ohm , i wonder where have i gone wrong
 
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thereddevils said:

Homework Statement



A 500 ohm resistor and a 1000 ohm resistor are connected in series with a 12 V battery of negligible internal resistance . When a voltmeter is connected across the 500 ohm resistor , its reading is 3 V . What is the resistance of the voltmeter ?


Homework Equations



V =IR

The Attempt at a Solution



the effective resistance across the voltmeter and 500 ohm resistor (i am not sure across which) is Re=500Rm/(500+Rm) where Rm= resistance of meter and Re=effective resistance

From V=IR , 12=I(1500) , so the current flowing through the circuit is not 0.008 A .

3=0.008(500Rm/(500+Rm))

solving this , Rm=1500 ohm

but the answer given is 1000 ohm , i wonder where have i gone wrong

1000 ohms is the correct answer. Your current flowing through the circuit is incorrect as you did not include the resistance of the ohmmeter in that calculation. The simplest way to solve it would be to determine the voltage across and then the current through your 1000 ohm resistor.
 
OmCheeto said:
1000 ohms is the correct answer. Your current flowing through the circuit is incorrect as you did not include the resistance of the ohmmeter in that calculation. The simplest way to solve it would be to determine the voltage across and then the current through your 1000 ohm resistor.

thank you