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    I know it's easy but i'm rubbish at circuits
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    (Original post by I'msoPi)
    I know it's easy but i'm rubbish at circuits
    The fuse ruptures when the current exceeds 5A. i.e. the maximum permissible current is 5A above which the fuse ruptures.

    The fuse is also a resistor. So the circuit is effectively two resistors in series connected to a 240V supply. i.e. a potential divider.

    Remember that the current at every place in a series circuit is the same.

    Use Ohms law to find the potential dropped across the lamp (V = I x R). Where I = 5A.

    Then remember Kirchoff's voltage law (KVL), that the sum of the voltages dropped in the series path, must equal the supply voltage.

    Vs = Vlamp + Vfuse

    We just worked out the voltage dropped across the lamp and we know the supply voltage, so the remainder must be the voltage dropped across the fuse.

    Finally, use ohms law again for the resistance of the fuse.

    Rlamp = Vlamp / I
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    (Original post by uberteknik)
    The fuse ruptures when the current exceeds 5A. i.e. the maximum permissible current is 5A above which the fuse ruptures.

    The fuse is also a resistor. So the circuit is effectively two resistors in series connected to a 240V supply. i.e. a potential divider.

    Remember that the current at every place in a series circuit is the same.

    Use Ohms law to find the potential dropped across the lamp (V = I x R). Where I = 5A.

    Then remember Kirchoff's voltage law (KVL), that the sum of the voltages dropped in the series path, must equal the supply voltage.

    Vs = Vlamp + Vfuse

    We just worked out the voltage dropped across the lamp and we know the supply voltage, so the remainder must be the voltage dropped across the fuse.

    Finally, use ohms law again for the resistance of the fuse.

    Rlamp = Vlamp / I
    Ah ha I understand now. Thank you so much, need to keep working on these circuits I think
 
 
 
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