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    First question asks to use a graph to determine the current in the resistor R. That was 0.8A which is correct.
    It then asks to calculate the resistance of the resistor R.
    \frac{8}{0.8}=10\Omega which is correct.
    The next part which I have no idea how to approach asks if the lamp Q were removed. Explain, without further calculation, how the voltmeter reading would change. The question carries three marks.
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    (Original post by halpme)
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    First question asks to use a graph to determine the current in the resistor R. That was 0.8A which is correct.
    It then asks to calculate the resistance of the resistor R.
    \frac{8}{0.8}=10\Omega which is correct.
    The next part which I have no idea how to approach asks if the lamp Q were removed. Explain, without further calculation, how the voltmeter reading would change. The question carries three marks.
    The lamp and P, which has some internal impedance, are in parallel. What will happen to the parallel impedance when you remove the lamp?
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    (Original post by atsruser)
    The lamp and P, which has some internal impedance, are in parallel. What will happen to the parallel impedance when you remove the lamp?
    The parallel impedance will be removed?
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    (Original post by halpme)
    The parallel impedance will be removed?
    That's correct, but in terms of the value of the impedance, which way must it go? Up or down?
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    (Original post by atsruser)
    That's correct, but in terms of the value of the impedance, which way must it go? Up or down?
    Up? That's a complete guess, can you explain why it must go up/down to me?
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    (Original post by halpme)
    Up? That's a complete guess, can you explain why it must go up/down to me?
    Up is right.

    Let's assume that P's impedance is purely resistive (call it R_1) and let the resistance of the bulb be R_2. Then we have for the total resistance:

    \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \Rightarrow R = \frac{R_1R_2}{R_1+R2} = \frac{R_1}{1+\frac{R_1}{R_2}} < R_1

    since  1+\frac{R_1}{R_2} > 1.

    So the total parallel resistance is less than the resistance of R_1 (and similarly it must be less than R_2).

    The total parallel resistance of two resistors is always less than the resistance of either. So if you remove one of the resistances from pair of parallel resistances, you can only increase the resistance.
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    (Original post by halpme)
    Name:  2013-08-14 20_15_27-6PH02 June 2011.jpg
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    First question asks to use a graph to determine the current in the resistor R. That was 0.8A which is correct.
    It then asks to calculate the resistance of the resistor R.
    \frac{8}{0.8}=10\Omega which is correct.
    The next part which I have no idea how to approach asks if the lamp Q were removed. Explain, without further calculation, how the voltmeter reading would change. The question carries three marks.
    The voltmeter reading will increase. Since you removed the lamp, the total resistance around the resistant P has increased.Hence, its share among the total circuit in series with R has increased.

    Rp=Resistance of P
    Rr=Resistance of R
    Vp=Voltage of P=Voltmeter reading

    Vp=Rp/(Rp+Rr) x 12 V

    Since Rp has gone higher, Vp will be higher.
 
 
 
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