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    The wires leading from the mains supply in the street into a building have a resistance R of 0.2 Ω. The electrical potential difference of the mains supply in the street is constant and equal to 230 V. The electrical potential difference at the sockets in the building is not to drop below 225 V when electrical power is consumed.

    i) Sketch a circuit diagram with the given figures noted on it.

    ii) Calculate the maximum current, and hence the maximum power that can be dissipated in the building so that the potential difference at the sockets does not drop below 225 V.


    In the attachment below, is the answers but I do not understand how to get that diagram from the information given.
    Also, I managed to get I = 25A but then to get the power, I dont see why they did 225 x 25 = P, I thought it would of been 25 x 5 (The drop in P.D.)...

    Thanks for any help
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    (Original post by soutioirsim)
    The wires leading from the mains supply in the street into a building have a resistance R of 0.2 Ω. The electrical potential difference of the mains supply in the street is constant and equal to 230 V. The electrical potential difference at the sockets in the building is not to drop below 225 V when electrical power is consumed.

    i) Sketch a circuit diagram with the given figures noted on it.

    ii) Calculate the maximum current, and hence the maximum power that can be dissipated in the building so that the potential difference at the sockets does not drop below 225 V.


    In the attachment below, is the answers but I do not understand how to get that diagram from the information given.
    Also, I managed to get I = 25A but then to get the power, I dont see why they did 225 x 25 = P, I thought it would of been 25 x 5 (The drop in P.D.)...

    Thanks for any help
    The diagram is easy enough, as it just shows a power supply of 230V a resistor and the PD across the second resistor, representing the circuit around the building. Why they have written 0.2 watts rather than ohms I don't know, only thing I can think of is they couldn't get an omega for some reason, or it's a typo.

    For part II the current is obvious, the power less so, the question doesn't actually make sense as if 5630W are dissipated then there would only be 120W left so the voltage would fall too much, I suspect the question is phrased badly.
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    I'm certain the W in the diagram is a typo. It should have been Ω for the resistance in the wires.
    The power available in the building is P=VI for the voltage, 225V, and current, 25A available in the building.
    They have rounded the answer up to 3 sig figs.
    25 x 5 would be the power lost in the o.2Ω resistor and therefore not available in the building.
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    I just find it hard to figure it out and stuff. I have a huge weakness for electricity.

    I mean they calculate the current in the wire but then use 225V for the power? Hmmmm...

    It is a British Physics A2 Olympiad Paper 1 question but it was one of the first/easy questions.
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    (Original post by soutioirsim)
    I just find it hard to figure it out and stuff. I have a huge weakness for electricity.

    I mean they calculate the current in the wire but then use 225V for the power? Hmmmm...

    It is a British Physics A2 Olympiad Paper 1 question but it was one of the first/easy questions.
    They use 225 for the power because power is volts times amps and it's 225V you have available in the house.

    You start out with 230V available outside, and lose 5V in the wiring.
    So the power available inside is the power you start with less the power lost in the wires. Current is 25A.
    Original power available = 230 x 25
    Power lost in wires = 5 x 25
    Power left over and available = 225 x 25
 
 
 
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