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    In the circuit, the batteries have negligible internal resistance and the voltmneter V has a very high resistance. What would be the reading of the voltmeter?Attachment 727152

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    (Original post by gannaspanner)
    In the circuit, the batteries have negligible internal resistance and the voltmneter V has a very high resistance. What would be the reading of the voltmeter?Attachment 727152

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    what do you mean?
    the picture is right there
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    attachment not found
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    (Original post by the bear)
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    the total resistance in the circuit is 60 + {40R/(40 + R)} ohms, where R is the resistance of the voltmeter.

    the voltage drop across the 60 ohm resistor must be 2.5 volts.

    you can find the current flowing through the 60 ohm resistor... then find R
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    8D(ii) i=(10- 8)/(40+10)=0,04A -> V voltometer = 8 + 0,04*40 = 9,6V
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    Sorry I meant this one
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    i=(10- 8)/(40+10)=0,04A -> V voltometer = 8 + 0,04*40 = 9,6V
 
 
 
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