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Why the current will be unaffected if C is increased?

  1. Dec 14, 2014 #1
    I am doing a multiple choice question and I can't understand why one of them is true.

    A capacitor of capacitor of capacitance C discharges through a resistor of resistance R. Which one of the following statements is not true?

    After checking the answer, it turned out that answer D which is ''After charging to the same voltage, the initial discharge current will be unaffected if C is increased'' is true, but I don't understand how.

    If C = Q/V = IT/V, then current would be proportional to current and so current would increase with it.

    Could someone explain why D is true, explaining what I got wrong when trying to solve it?

    Thank you physicians.
     
  2. jcsd
  3. Dec 14, 2014 #2
    That formula is not right, not here.
    You can write Q=IT for a finite time interval if the current were constant.
    This is not the case here. After all they ask about "initial discharge current", so the initial value.

    Do you know the formula for discharge current? Have you studied this topic?
    If not, you can still find it if you consider that at the initial moment the initial voltage is applied on the resistor.
    Indeed the initial value does not depend on the capacitance. How fast will that current decay does.
     
  4. Dec 14, 2014 #3

    Dale

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    Q=IT is not generally correct. A better formula would be ##\Delta Q=I\Delta T##.

    Does that help you understand the problem?
     
  5. Dec 14, 2014 #4
    But not here. The current is a function of t so you have to integrate over t. Unless by "I" you mean the average current.
    But the question is about an instantaneous value of the current, not an average.
     
  6. Dec 15, 2014 #5

    CWatters

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    Should be more like..

    C=Q/V
    Q = VC
    I = dQ/dT = CdV/dT

    but that it doesn't really help solve your problem.

    The initial current is the initial voltage divided by the resistance. The initial voltage is fixed because the d) says "After charging to the same voltage...."
     
  7. Dec 15, 2014 #6

    Dale

    Staff: Mentor

    So, since using the formula for the capacitor is causing problems, perhaps look at using the formula for the resistor.
     
  8. Dec 15, 2014 #7
    The equations that help is
    IR=V=Q/C
    where you use I=dQ/dt
    This will give a differential equation in Q
    dQ/dt=Q/(RC)
    and from here you can find Q(t) and then I(t).
     
  9. Dec 15, 2014 #8

    sophiecentaur

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    [QUOTE="Tangeton, post: 4945849, member: 521367"''After charging to the same voltage, the initial discharge current will be unaffected if C is increased'' is true, but I don't understand how.

    [/QUOTE]
    The current just depends only upon the value of the Resistor and the Volts across it. The total Charge, depends upon the Capacitor value and the time taken for the volts to drop to a given value will be proportional to the R times the C (the time constant). The message in answer D is similar to saying "what is the current through a 1000 Ohm resistor when connected to a 1.5V AAA battery, and what is the current when connected to a massive 1.5V D type Cell? "
    A good tip for working out any circuit problem is to only consider what is actually necessary to solve it. You can ignore the C value, in this case. (And you have to assume that the Capacitors are ideal - with no series resistance)
     
  10. Dec 18, 2014 #9
    Please no differentiation.. I never learned how to differentiate anything in physics.

    I don't know much about this topic... My teacher doesn't explain things very well and we went through this topic like a tractor racing F1 cars, barely making a whole round. All do is try to look on related equations in formula book. I am given c = Q/V and Q = Q0e0t/RC for capacitors. I don't know how to use the second one very well. Can the second formula be used to solve this problem? If so, how and why?

    If not then I may as well just leave it unanswered...
     
  11. Dec 18, 2014 #10

    sophiecentaur

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    That's an unreasonable stipulation - like expecting an answer that doesn't contain the words "Charge" and "Current". You have to be prepared to learn the language if you want to get on in Science. If you are not happy with your teacher then you need to read round a lot. There are countless sites at all levels that can help you with Calculus - at a level that suits you. You cannot afford to reject the use of Maths just because you haven't been taught it in a way that appeals to you (not if you want to learn any useful Physics).

    dQ/dt just stands for 'the rate at which the charge changes with time'. A pretty good abbreviation, don't you think?
     
  12. Dec 18, 2014 #11
    I am prepared to learn the language but what I tried to say is that if I haven't been thought it I don't think I will need it for this question... I want to see if there is a way to do it without the use of differentiation, that's all.
     
  13. Dec 18, 2014 #12

    sophiecentaur

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    I can sympathise to some extent with that but to my mind, it's a bit like learning to use a potato printing kit when you have been promised a Word Processor in the near future. You don't need to know 'all' Calculus to do the simple stuff here. In this case, how I described the relationship dQ/dt is all you really need. It so happens that, in the case of a discharging capacitor, the way the current changes follows a fairly simple relationship
    I= I0 exp(-t/RC)
    I0 is the initial current (set by V/R) and I stands for the current at any subsequent time t. (Exponential decay)

    But the point of your original question is that the initial current is not affected by the value of C. The above stuff about charge rates etc. will probably come later. But if you are not happy with your teacher, it may be just as well to be ready for the next steps or you risk a few wasted lessons whilst you catch up with what he/she is introducing. You will have a Subject Specification available from somewhere, so you can see what you are likely to be getting in the future. In the UK, the Exam Boards all publish this info and I'm sure that the same is available everywhere. Ask the teacher (it's not being cheeky).
     
  14. Dec 18, 2014 #13
    I do read around but I didn't find anything similar in my books that's why I asked on here. I just simply didn't find any clue as to how to approach this question, and I didn't see any rates of change being included in my physics book, I just barely covered rates of change in maths.
     
  15. Dec 18, 2014 #14

    sophiecentaur

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    The secret for solving most / all circuit problems is to look at what you know and what you need to know. You will always (at your level) find one or two equations that will contain what you need to know. If one of them has terms in it that you have values for then you have your answer directly.

    Sometimes it isn't as simple as that and then you can look and see what the 'known' quantities can give you (another equation). This will almost always provide a link (by way of just one extra step) and you can fill in the first equation and you've done it. Seriously, the problems are always presented in a way that becomes very familiar and you can soon spot what's required from a mile off. (With practice and it's a lot easier than an adventure game on the computer.) Have you done the SUVAT equations in Mechanics, yet? It's a similar thing.

    As life gets harder, you can find that you end up with a number of simultaneous equations to solve (not too hard) and there are methods for reducing that sort of problem to give you the answers.
     
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