Determining current from charge function

1. Jun 11, 2008

scholio

1. The problem statement, all variables and given/known data

the amount of charge flwoing past a point in a current in a circuit is found to vary with time as follows:

Q(t) = 100[e^(+t/100) + 0.01t^2]

what is the current flowing at t = 10 seconds?

i am supposed to get 47.2 amperes

2. Relevant equations

C = Q/V ---> Q = CV where C is capacitance, V is electric potential, Q is charge

V = IR ----> I = V/R = dW/dt where R is resistance, W is work, t is time

3. The attempt at a solution

first solve the function in terms of t = 10 seconds

Q(10) = 100[e^(+10/100) + 0.01(10^2)] = 100[ 1.105 + 1] = 100(2.105) = 210.5 coulombs

since Q = CV and V = IR ---> Q = CIR now if i let Q = 210.5 coulombs, i don't have values for R or C, i'm solving for I.

am i using the wrong approach entirely or the wrong equation?

help appreciated

Last edited: Jun 11, 2008
2. Jun 11, 2008

rock.freak667

The current flowing is numerically equal to the The rate of change of charge per unit time

3. Jun 11, 2008

scholio

do you mean I = dQ/t? i thought I = dW/dt, anyways if i used I = dQ/t = 210.5/10 = 21.05 amperes, i'm supposed to get 47.2 amperes

by the way, did i solve for charge Q correctly?

4. Jun 11, 2008

nicksauce

I = dQ/dt... meaning you have to differentiate Q(t).
I(10) = dQ/dt | t=10, not I = Q(10)/10... they are completely different expressions.

5. Jun 11, 2008

rock.freak667

$$\frac{dW}{dt}=Power$$

but I think you'll get the 21.05A. Not sure how to squeeze the 47.2A answer though.
But that is how you would do the problem,using $I=\frac{dQ}{dt}$

6. Jun 11, 2008

scholio

you are right, dW/dt is power, i must've gotten it confused with dQ/dt, thanks for clearing that up.

did i solve for Q correctly though?

7. Jun 11, 2008

rock.freak667

Looks correct to me.

8. Jun 11, 2008

scholio

does anybody know why i am getting 21.05 ampere, when i should be getting 47.2 ampere?

thanks