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    Hey can anyone help em out with this question? I've attempted it but I'm pretty sure that I got . it wrong and I don't know hat the correct answer is.

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    it gave a IV graph for filament lamp and a resistor before this and asked to calculate the resistance of the resistor for which I got 14.5Ohms.
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    (Original post by FaNtAsY FlAmE 7*)
    it gave a IV graph for filament lamp and a resistor before this and asked to calculate the resistance of the resistor for which I got 14.5Ohms.
    Scan and post the graph.
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    Here they are. I've only managed to do the first one and I think the other link to one another but I'm not sure.
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    (Original post by FaNtAsY FlAmE 7*)
    Here they are. I've only managed to do the first one and I think the other link to one another but I'm not sure.
    Both components are connected in series which means the current is the same at all points in the series path.

    Therefore 0.18A flows through both the resistor and the lamp.

    The resistor is linear as shown by the straight line IV graph and as you correctly deduced, is around 14.5 ohms.

    Which means the voltage drop across the resistor for 0.18A current is:

    V_{R} =  I \text{ x }R = 0.18 \text{ x } 14.5 = 2.6 \text{volts}

    Again reading from the graph, this time for the lamp characteristic, the voltage required to produce a current flow of 0.18A through the lamp is around 0.8 volts.

    From Kirchoff's voltage rule, the sum of the voltage drops in a series path must equal the supply voltage.

    Vsupply = VR + Vlamp

    Vsupply = 2.6 + 0.8 = 3.4 Volts.
 
 
 
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