Finding 'J' over 's' with a voltage drop?

Discussion in 'Homework Help' started by Lolpz, Sep 7, 2010.

  1. Lolpz

    Thread Starter New Member

    Sep 7, 2010
    Hi everybody, I am a freshman college student who is greatly interested in electrical circuits for my major in hardware engineering but I am struggling on my first homework assignment for my ELEC 1300 class.

    This is my first post but since I have found out about this forum, I hope to continue posting and contribute back to the community as I learn more, in return for the help others can offer me.

    I only say this because I don't want the community to think I'm just trying to get some quick answers to get me by in the class with no work done on my own part.

    I would like to included a short bit of background on myself to help you help me. The highest form of math I have taken as of yet is Pre-calculus my senior year in high school. When choosing my courses for college, the prerequisite for this electric circuits class was only pre-calculus so I thought I would be fine, however once I started, the professor told us we would be using calculus from day 1. I'm still slightly confused about integrals and know that I need to understand them to solve many of the problems. I say this because if integrals are involved in either of these questions, I would greatly appreciate if you could explain why and how you used integrals to solve.

    With that said, here are my two questions (They are relatively unrelated so I only posted one in the subject line)

    The first question I have, I did a search and found a forum already posted with a very similar question, however, I was confused by the answer provided and was wondering if somebody could help explain better. Here is the link if you would like to check it out first:

    1.) The current entering the upper terminal of Fig. 1.5. (Included in pdf and jpeg as attachment) is "i = 20cos5000t A."

    Assume the charge at the upper terminal is zero at the instant the current is passing through its maximum value. Find the expression for q(t).

    I'll be honest and say I don't understand anything about this question. I have been paying attention during class and am positive the professor hasn't mentioned finding expressions for q(t) and cos was never involved in any work we did.

    I know I'm suppose to show my attempts so my only guess as to what I'm trying to do here is solve for 'q', since I am given 'i' and 't' and I know "q = it", however, 't' is already included in current given so I am completely lost. Basically, I don't know what it means when it says "Find the expression for q(t)". Sorry I can't provide any other knowledge about the problem but this just left me completely confused. Also, the answer in the back of the book is "4sin5000t mC" but I have no idea how this answer could be reached.

    My second problem I understand a bit more and could take a guess but am near positive I would be wrong.

    2.) The manufacturer of a 1.5 V D flashlight battery says that the battery will deliver 9 mA for 40 continuous hours. During that time the voltage will drop from 1.5 V to 1.0 V. Assume the drop in voltage is linear with time.

    How much energy does the battery deliver in this 40 h interval?

    I'm trying to solve for 'J' (energy) and I am given 'V', 'T', and "i". Thus, I assume I can use the power formula P = (w/t) = VI.

    First I convert my units, so
    9mA = .009A
    40h = 144000s
    Also, if V drops from 1.5 to 1.0 over time, I assume I should use 1.25 as V?

    Solving for P I get (1.25)x(.009) = .01125W

    Now using P = (w/t) I get .01125 = (w/144000) which comes out to w=1620J

    Thank you very much in advance for any help you can offer me and sorry about the excessive text at the top. Hopefully it shows that I'm really interested in being part of the community and am not just here this once begging for answers.
    Last edited: Sep 8, 2010
  2. Georacer


    Nov 25, 2009
    Hello Lolpz and welcome to the AllAboutCircuits community

    From day 1, you proved your will to try and made noticeable efforts before asking for help, so don't worry, we'll take it from here.

    For question 1:

    First off a clarification. The general equation that relates current and charge is  i=\frac{dq}{dt}. If the current is constant, then it translates toq=i\cdot t. But in the general situation that the current is arbitrary and non-constant, the current that flows through a conductor from time t1 until time t2 is  q=\int\limits_{t_1}^{t_2} i(t) dt, wich, yes, to your dismay, is an integral. But fret not! if you don't understand a step, feel free to ask.

    Now, we need to find the charge that flowed through the upper pin from time 0 until the (arbitrary) time "t". So we set t1=0 and t2=t. Let's write it again:
     q=\int\limits_{0}^{t} i(t) dt
    A note of warning: It is important to notice that we indeed wanted to start from time 0, because that is the moment that the current is at its maximum (since it is a cosine waveform). If we had a sine in our hands we should start a quarter of a period later. So we have:
     q=\int\limits_{0}^{t} 20 \cos (5000 \cdot t) dt\\<br />
\Leftrightarrow q= \int\limits_{0}^{t} 20 \cdot \left(\frac{\sin(5000 \cdot t)}{5000}\right)' dt\\<br />
\Leftrightarrow q=20 \cdot \left( \frac{\sin (5000 \cdot t)}{5000} \right) -q(0)\\<br />
\Leftrightarrow q=0.004 \cdot \sin (5000 \cdot t) \text{ in Coulombs, or}\\<br />
\Leftrightarrow q=4 \cdot \sin (5000 \cdot t)\  mC<br />

    Note that q(0)=0, as the problem states. That's about it! Any questions?

    For question 2:
    I don't find anything wrong, well done! You could use an integral to calculate Power as a function of time, as Voltage is time-dependant, but since the relation is linear, the little "trick" you did, evaluating it as 1.25 Volts, is valid.
  3. Lolpz

    Thread Starter New Member

    Sep 7, 2010
    Awesome! Thank you so much, I didn't even understand what the question was asking but I went through each step you provided and understand it much better now.

    The only part that I did not understand fully was when you subtracted q(0) in the 3rd to last step.

    I know q(0) = 0 so it did not matter in this particular case, but just in case I solve another similar problem in which q(0) has a value, could you explain why you subtracted q(0)? In the step above, there was no subtraction at the end of the equation, so when you subtracted q(0) I became a little lost.

    Also, I'm surprised I got number 2 right, but since it said it was linear I guess I should have known this had to be correct.

    Thanks again for the help! Heading off to class now, have a nice day :)
  4. Lolpz

    Thread Starter New Member

    Sep 7, 2010
    Oh wait, I believe I may have figured out why you subtracted q(0). Was it because to get the charge during the specified time (from the instant it is connected -> t), you have to subtract the charge at t=0, from the charge at time t? Was just waiting for class to start and thought about that heh. Anyways, thanks again.
    Last edited: Sep 8, 2010
  5. Georacer


    Nov 25, 2009
    A quick integration overview:

    Say we have a function f(x) and a function F(x) for wich it is valid that f(x)=F(x)'. That means that f(x) is the derivative of F(x), or F(x) is the antiderivative of f(x).

    The way to get f(x) from F(x) is the derivation: As we said before f(x)=F(x)'.

    The way to get F(x) from f(x) is the integration. F(x)=\int f(x) dx.
    The use of integration is to allow us to reconstruct F(x) if we only know f(x). f(x) tells us how F(x) slopes, how it increases or decreases, as the independent variable x proceeds on the horizontal axis. What f(x) cannot tell us, is how high in the y-axis is F(x) placed. It can tell us if it goes up or down, but it can't tell us how high or how low it actually is in the cartesian plane.

    Say now we have a function G(x)=F(x)+c, where c is an arbitrary constant. For G(x) it is also valid that f(x)=G(x)'=F(x)'+c'=F(x)'. That means that G(x) is also an antiderivative of f(x). We come to the conclusion that every function of the form F(x)+c is an antiderivative of f(x). Actually, the antiderivative of f(x) is a familly of functions with infinite members. This constant c is exactly that piece of information that f(x) couldn't tell us in the previous paragraph.

    So if we try to integrate f(x) from point a and on, the answer we will seek is a function, say H(x) for wich we have:
    H(x)=\int\limits_a^x f(x) dx\\<br />
\Leftrightarrow H(x)=F(x)-F(a)
    By inputing the value of F(x) at the beginning of the space we need evaluated (wich is F(a)), we tell the integration the initial value of the funtion we seek. It is the same as defining the constant c, wich chooses one function out of the whole familly

    Back to our problem, by knowing the current flowing in the pin, we knew how the charge changed. But we wouldn't be able to know the actual value of the charge, unless we had a point of referrence. This is exactly what the initial condition is. A starting value, zero for our situation, based on wich we will know where to begin, when we say that, i.e. the charge increased by 2 Cb. We can conclude that the charge is now 2 Cb. If the initial value was -1 Cb, it present state would be 1 Cb.

    I blabbered too much, but I hope I managed to clarify the importance of the initial condition.

    edit: You posted the post above while I was writing this reply. Yes you got the general idea but read the text above for a more in-depth analysis.
  6. jercie

    New Member

    Jan 28, 2016
    how did you get V=1.25?