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:
- \(c - 3 = 3\)
\[ c = 3 + 3 = 6 \]
- \(c - 3 = -3\)
\[ c = 3 - 3 = 0 \]
Thus, the currents that will produce no power are \(0\) amperes and \(6\) amperes.
To summarize:
Lower current: \(0\) amperes
Higher current: \(6\) amperes