   Chapter 2.7, Problem 19E

Chapter
Section
Textbook Problem

# The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given by Q ( t )   =   t 3 −  2 t 2 +   6 t   +   2 . Find the current when(a) t = 0.5 s and (b) t = 1 s. [See Example 3. The unit of current is an ampere (1 A = 1 C/s).] At what time is the current lowest?

To determine

To find: the current when

(a) t=0.5 s

(b) t=1 s

Explanation

1) Formula:

Power rule,

ddxxn=n*xn-1

2) Given:

Qt=t3-2t2+ 6t+2

3) Calculation:

Find derivative of given function to obtain an equation for current,

Q't=

To determine

(a) The current at t=0.5 s is 4.75 A

(b) The current at t=1 s is 5 A

To find: time when current is lowest.

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