Physics maths calculus charge calculations

  • #1

Homework Statement


charge (q) in coulombs, can be calculated byt integrating the current (i) in amperes (A) with respect to time. determine the charge on a capacitor as a result of current flowing for 0.03 seconds if:
a. the current is a constant 5mA (0.005A)
b. the current is a function of time given by i=60t^2


Homework Equations


i believe q=(i)(t)


The Attempt at a Solution


if current (i) is integrated with respect to time = it^0+1/0+1 = q=i*t^1 = q=(i)(t)
a. i believe to be q=0.005*0.03=0.00015 coulombs

b. however i assume that i need to integrate i=60t^2
but the A/mA is confusing me i think

should it be 20t^3 or 0.02t^3?

and to find the charge do i just multipli the answer by t ? or is there any further calculation needed?

any help would be appreciated

i

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
108
0
I would integrate the function as it's given to you.

(1) [tex] \int_{0}^{0.03} 0.005 dt [/tex]

(2) [tex] \int_{0}^{0.03} 60t^2 dt [/tex]

For (1), it becomes clear by integration that it is indeed just I*t, the integration of a constant function multiplied by the elapsed time.

For (2), I think you're on the right track. I wouldn't amend 60 to be 0.06 since the whole functional expression (60t2) is your current. It would be useful if they let you know if that expression is A or mA, but since there is no information otherwise, I would assume it to be A. You can also compare your answers for the two questions if that makes it at all clearer.
 
  • #3
sorry i forgot to add that to the question part b. is to calculate current in mA
 
  • #4
108
0
sorry i forgot to add that to the question part b. is to calculate current in mA

So that functional expression is in mA? In that case, I'd still do the integration like normal, just convert afterward. (you integrate in mA, so you have mA*s = mC...now convert to C)
 
  • #5
so i should work with 20t^3

(20*0.03^3)-(20*0^3)=0.00054-0=0.00054mA

=0.0000054A

q=0.0000054*0.03=0.000000162coulombs

thank you for your comments
 

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