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Physics, Azzopardi book page 291 question 3 watch

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    please help I cannot find out how to answer this and it is very worrying
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    Sorry you've not had any responses about this. Are you sure you've posted in the right place? Here's a link to our subject forum which should help get you more responses if you post there.


    Just quoting in Fox Corner so she can move the thread if needed :wizard:
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    Could you please post the question?
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    (Original post by tempo91)
    Could you please post the question?
    Two 15 micro-Farad capacitors are connected in parallel, and the pair are connected in series with a 45 micro-Farad capacitor and a battery. If the p.d. across the parallel pair is 30 V, Calculate:
    a) the charge on each capacitor
    b) the e.m.f. of the battery
    c) the total capacitance of the circuit
    d) the total energy stored in the capacitors.
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    Firstly it would help to draw a circuit a diagram to understand the scenario.

    For now, let the two 15 micro-Farad capacitors and 30 micro-Farad capacitor be Q_1, Q_2 and Q_3 respectively.

    As for part a, you can calculate the charge on the capacitors in parallel because the question tells you the potential difference and capacitance across the parallel components using Q = VC. To calculate the charge on 45 micro-Farad capacitor, you can treat the parallel components as if they were a single component equivalent to the ones in parallel. So since charge and capacitance can be added in parallel, so you'd have a 30 micro-Farad (15 + 15) and a 45 micro-Farad capacitor bearing in mind the charge on the 30 micro-Farad capacitor is now the sum of charge across the 15 micro-Farad capacitors.

    You know in series, the charge across all capacitors are the same so Q_1 + Q_2 = Q_3

    As for part b, the e.m.f of the battery is the sum of all potential difference across the components. You know that 30V goes across the capacitors in parallel and you can use V = \frac{Q}{C} to work out the p.d. across the 45 micro-Farad capacitor. Add everything up and you'll get your answer.

    As for part C, do the same method from part a and treat the circuit as if it was a series circuit.

    As for part d, use an energy equation where your C is the total capacitance and V is your e.m.f. A long method is to calculate the energy on each capacitor and then add them up.
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    (Original post by tempo91)
    Firstly it would help to draw a circuit a diagram to understand the scenario.

    For now, let the two 15 micro-Farad capacitors and 30 micro-Farad capacitor be Q_1, Q_2 and Q_3 respectively.

    As for part a, you can calculate the charge on the capacitors in parallel because the question tells you the potential difference and capacitance across the parallel components using Q = VC. To calculate the charge on 45 micro-Farad capacitor, you can treat the parallel components as if they were a single component equivalent to the ones in parallel. So since charge and capacitance can be added in parallel, so you'd have a 30 micro-Farad (15 + 15) and a 45 micro-Farad capacitor bearing in mind the charge on the 30 micro-Farad capacitor is now the sum of charge across the 15 micro-Farad capacitors.

    You know in series, the charge across all capacitors are the same so Q_1 + Q_2 = Q_3

    As for part b, the e.m.f of the battery is the sum of all potential difference across the components. You know that 30V goes across the capacitors in parallel and you can use V = \frac{Q}{C} to work out the p.d. across the 45 micro-Farad capacitor. Add everything up and you'll get your answer.

    As for part C, do the same method from part a and treat the circuit as if it was a series circuit.

    As for part d, use an energy equation where your C is the total capacitance and V is your e.m.f. A long method is to calculate the energy on each capacitor and then add them up.
    Correct . figured it out the next day and i forgot about this post but your method is correct D
 
 
 
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