#### To determine

**To find:** The magnitude of current when the value of time is given as t=0.5 s, also the time at which the lowest value of current occurs.

#### Answer

Therefore, the current when time t is equal to 0.5s is 4.75A.

#### Explanation

**Given:**

The quantity of charge Q in coulombs (C) that has passed through a point in wire up to time t is given by the equation

Q(t)=t3−2t2+6t+2 (1)

The unit of current is in amperes (A).

**Calculation:**

Differentiate the equation (1) as below.

Q'(t)=3t2−4t+6

Calculate the current when t is 0.5 s.

Substitute the value of 0.5 s for t in above Q'(t) equation.

Q'(0.5)=3(0.5)2−4(0.5)+6Q'(0.5)=4.75A

Therefore, the magnitude of current when time t is equal to 0.5 s is 4.75 A_.

#### To determine

**To find:** The magnitude of current when the value of time is given as t=1 s, also the time at which the lowest value of current occurs.

#### Answer

The current when time t is equal to 1s is 5A_.

The current is lowest at the time equal to 23sec_ .

#### Explanation

Calculate the current when time t is equal to 1 s.

Substitute the value 1 s for t in above Q'(t) equation.

Q'(1)=3(1)2−4(1)+6Q'(1)=5A

Therefore, the current when t is equal to 1 s is 5 A_.

Calculate at what time the current is lowest.

The current is lowest when the Q' has a minimum value.

Differentiate Q' again,

Q"(t)=6t−4

Equating the Q"(t) to zero.

Q"(t)=6t−4=06t−4=06t=4t=46t=23s

When the value of time t<23 s, current is decreasing.

Substituting the value of 12 s for t in Q''(t).

Q''(12)=6(12)−4Q''(12)=−1A < 0

When the value of time t>23, current is increasing.

Substituting the value 1 for t in Q''(t).

Q''(1)=6(1)−4Q''(1)=2A > 0

Therefore, the current is lowest at the time 23sec.