Electricity Quiz

    Question One

    The circuit shown has four identical ammeters. A2reads 0.2A and A3reads 0.3A


    What would the current be through the other two ammeters?

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    They would both read the same value - and that would be 0.5A.

    The sum of the 2 currents in the parallel pathways is equal to the current before it split (0.5A). When the pathways rejoin the current is again 0.5A. Current is NOT used up around a circuit.

    Question Two

    Which one of the following statements is correct?

    • The resistance of P is more than 20 ohms.
    • The resistance of P is equal to 20 ohms.
    • The resistance of P is less than 20 ohms.

    Give a reason for your choice.

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    The resistance of P is more than 20 ohms.

    Since less current flows through A2 than A3, P has more resistance than 20 ohms.

    (If they had been the same resistance then the current through each ammeter would have been the same. If P had had a lower resistance than 20 ohms then more current would have flowed through that pathway).

    Question Three

    Write down the equation that links current, potential difference and resistance.

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    V=IR where V is voltage or potential difference, I is current and R is resistance.

    Question Four

    Calculate the reading on the voltmeter. Show clearly how you work out your answer.

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    I=0.3A, R = 20 ohms

    Since V=IR then V=0.3 x 20 = 6V

    Question Five

    State the potential difference of the power supply.

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    This would also be 6V.

    Question Six

    Calculate the resistance of resistor P. Show your working.

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    The voltage across P would also be 6V. The current through P is 0.2A.

    V=IR so R=V/I. R=6/0.2 = 30 ohms.

    (Look back at Q2. yes we said the resistance of P was > 20 ohms!)

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