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Math Help - how much charge is transported.

  1. #1
    Super Member bigwave's Avatar
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    Cool how much charge is transported.

    a constant current of 3A for 4 hours is required to charge an automotive battery. If the terminal voltage is 10 + t/2\ V, where t in hours

    (a) how much charge is transported as a result of the charging?

    (b) how much energy is expended?

    (c) the much does the charging cost? Assume electricity costs 9 cents/kWh

    first was wondering the why the expression 10 + t/2\ V is used

    thanks again for help
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  2. #2
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    Quote Originally Posted by bigwave View Post
    a constant current of 3A for 4 hours is required to charge an automotive battery. If the terminal voltage is 10 + t/2\ V, where t in hours

    (a) how much charge is transported as a result of the charging?

    (b) how much energy is expended?

    (c) the much does the charging cost? Assume electricity costs 9 cents/kWh

    first was wondering the why the expression 10 + t/2\ V is used

    thanks again for help
    Dear bigwave,

    For a), use the definition of current. I=\frac{dq}{dt} where I is the current, q is the charge and t is the time. In this case the current is constant. Therefore this gives, q=It

    For, b) you should know that the energy expended is given by, E=VIt where V is the terminal volatage. Hope you know have a idea why the terminal voltage expression is given.

    For c), first find out the kilowatt hours that had been used because of the charging. The power is given by, P=VI. You will have to multiply this with the hours used. Then use the given cost per kWh to estimate the charging cost.

    Hope you will be able to continue.
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  3. #3
    Super Member bigwave's Avatar
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    Cool Q=43.2kC

    well for (a) using \int_0^{14400}(3)t=43200or 43.2k[C]

    which is the books ans. however don't see need for the intergal

    but for (b) the books ans is 475K[J] but (10+\frac{14400}{2})(3)(14400)<br /> <br />
=3.1147 which is not the ans

    actually don't know why the voltage expression 10+\frac{t}{2}[V] is used
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  4. #4
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    Quote Originally Posted by bigwave View Post
    well for (a) using \int_0^{14400}(3)t=43200or 43.2k[C]

    which is the books ans. however don't see need for the integral
    Because \displaystyle dq=I\cdot dt\ \to\ q=\int_0^{14400}(3)\,dt

    Quote Originally Posted by bigwave View Post
    but for (b) the books ans is 475K[J] but (10+\frac{14400}{2})(3)(14400)<br /> <br />
=3.1147 ×10^8 which is not the ans

    actually don't know why the voltage expression 10+\frac{t}{2}[V] is used

    It's given because: \displaystyle Power={{dE}\over{dt}}=I\cdot V

    This gives: \displaystyle E=\int_0^{14400} I\cdot V \, dt = \int_0^{14400}(3)\cdot \left(10+{{t}\over{2}}\right) \, dt\,. Of course this will be in Joules.

    For (c) use the fact that 1 Joule = 1 Watt·Second.

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    Mathematics is the study of quantity, structure, space, and change. Mathematicians seek out patterns, formulate new conjectures, and establish truth by rigorous deduction from appropriately chosen axioms and definitions. For more details Search in Wikipedia.
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    Post # 5 here is totally irrelevant to this thread. Could a mod please delete?
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  7. #7
    Super Member bigwave's Avatar
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    Cool where the "10+" came from

    Quote Originally Posted by SammyS View Post


    This gives: \displaystyle E=\int_0^{14400} I\cdot V \, dt = \int_0^{14400}(3)\cdot \left(10+{{t}\over{2}}\right) \, dt\,. Of course this will be in Joules.

    I was curious where the "10+" came from in the voltage
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  8. #8
    Super Member bigwave's Avatar
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    475.2kJ

    Quote Originally Posted by SammyS View Post

    This gives: \displaystyle E=\int_0^{14400} I\cdot V \, dt = \int_0^{14400}(3)\cdot \left(10+{{t}\over{2}}\right) \, dt\,. Of course this will be in Joules.
    this did not give me the answer of 475.2kJ
    which is the answer in the book...
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  9. #9
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    Hi Bigwave

    \displaystyle\ 3\int_{0}^4{\left(10+\frac{t}{2}\right)}dt=3\left(  10[4]+\frac{4^2}{4}\right)=132\;\;Watt-hours

    1Wh=3.6kJ

    132(3.6)=475.2kJ

    Also, the expression for battery voltage...
    Voltage at beginning of charge=10V.
    Voltage ramps up linearly for 4 hours to 12V.
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