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    The power supplies are the same. The bulbs have the same resistance and the ammeter has negligible internal resistance.

    By what percentage does the ammeter reading, increase or decrease when the switch is closed? Or does it remain the same.

    Please explain why.

    Thanks very much,

    Vazzyb. Rep available!
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    Ammeter reading will drop to zero.

    If you apply Kirchoff 2 round the outside loop.

    Sigma emf = 0
    so PD across bulb =0
    so current through ammeter =0
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    Very sorry about that, i meant to put the second source the other way around. I have corrected the circuit. Repped you anyway.
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    Current now goes from E/3R to 2E/R. (Apply Kirchoff 2 around outside loop)
 
 
 
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