- #1

electro 1982

- 3

- 0

## 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