The power generated by an electrical circuit (in watts) as a function of its current

Question

\[c\] (in amperes) is modeled by:
\[P(c)=-20(c-3)^2+180\]
Which currents will produce no power (i.e.
\[0\] watts)?
Enter the lower current first.
Lower current:
amperes
Higher current:
amperes

Answer 1

To find the currents that produce no power (P(c) = 0), we need to solve the equation:

\[ -20(c-3)^2 + 180 = 0 \]

First, we can isolate the squared term:

\[ -20(c-3)^2 = -180 \]

Dividing both sides by -20 gives:

\[ (c-3)^2 = 9 \]

Now we take the square root of both sides:

\[ c-3 = 3 \quad \text{or} \quad c-3 = -3 \]

Solving for \(c\) in both cases:

1. \(c - 3 = 3\)

\[ c = 3 + 3 = 6 \]

2. \(c - 3 = -3\)

\[ c = 3 - 3 = 0 \]

Thus, the currents that will produce no power are \(0\) amperes and \(6\) amperes.

To summarize:

Lower current: \(0\) amperes
Higher current: \(6\) amperes