Question

HELP PLEASE!!!!!!!!!!!!!!!!




The power generated by an electrical circuit (in watts) as a function of its current ccc (in amperes) is modeled by:

P(c)=−20(c−3)^2 + 180

Which currents will produce no power (i.e. 000 watts)?
Enter the lower current first.

Lower current: _______ amperes
Higher current: _______amperes

Answer 1

Answer:

Lowest Current : c=0 and 6 Amp

Highest Current : 3 amp

Step-by-step explanation:

We are given our function as

$P(c)=-20(c-3)^2 + 180$

We are asked to determine the values of current c at which the power P(c) is equal to 0

Hence

$0=-20(c-3)^2+ 180$

Now we solve the above equation for c

subtracting 180 from each side we get

$-180=-20(c-3)^2$

Dividing both sides by -20

$(c-3)^2=9$

Taking square root on both sides

c-3= ±3

adding 3 on both sides

c=±3+3

hence

c= 0

or

c=6

At c=0 and 6 amperes the power will be minimum

Now we have to find the c at which the power will be the highest

$P(c)=-20(c-3)^2+ 180$

Represents a parabola

subtracting 180 from both sides we get

$P-180=-20(c-3)^2$

Comparing it with standard parabola

$(y-k)^2=-4k(x-h)^2$

(h,k) will be the coordinates of the vertex

Hence here

h=3 , k = 180

Hence in this equation $P-180=-20(c-3)^2$

The vertex will be (3,180)

Or at c=3, P = 180 the maximum

Answer 2

Answer:

Lower current: 0 amperes  

Higher current: 6 amperes

Step-by-step explanation:

Hi, to answer this question we have to replace the value of P by 0 and solve.

0 = -20 (c-3)² +180

-180 = -20 (c-3)²

-180/-20 =(c-3)²

9=(c-3)²

√9 = c-3

±3= c-3

• For the positive option

3+3= c

6 =c

• For the negative option

-3 =c-3

-3+3 =c

c=o

Lower current: 0 amperes  

Higher current: 6 amperes

Feel free to ask for more if needed or if you did not understand something.