# 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