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

The power generated by an electrical circuit (in watts) as a function of its current c c (in amperes) is modeled by:\newlineP(c)=−20(c−3)2+180 P(c)=-20(c-3)^{2}+180 \newlineWhich currents will produce no power (i.e. 0 0 watts)?\newlineEnter the lower current first.\newlineLower current: amperes \text{amperes} \newlineHigher current: amperes \text{amperes}

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Q. The power generated by an electrical circuit (in watts) as a function of its current c c (in amperes) is modeled by:\newlineP(c)=−20(c−3)2+180 P(c)=-20(c-3)^{2}+180 \newlineWhich currents will produce no power (i.e. 0 0 watts)?\newlineEnter the lower current first.\newlineLower current: amperes \text{amperes} \newlineHigher current: amperes \text{amperes}
  1. Given power function: We are given the power function P(c)=−20(c−3)2+180P(c) = -20(c - 3)^2 + 180. To find the currents that produce no power, we need to solve the equation P(c)=0P(c) = 0.
  2. Set equal and solve: Set the power function equal to zero and solve for cc:0=−20(c−3)2+1800 = -20(c - 3)^2 + 180
  3. Move 180180 and solve: Move 180180 to the other side of the equation:\newline−20(c−3)2=−180-20(c - 3)^2 = -180
  4. Divide and isolate: Divide both sides by −20-20 to isolate the squared term:\newline(c−3)2=9(c - 3)^2 = 9
  5. Take square root: Take the square root of both sides to solve for cc:c−3=±3c - 3 = \pm 3
  6. Solve for c: Solve for c by adding 33 to both possible values of the square root:\newlinec=3±3c = 3 \pm 3
  7. Two solutions for c: This gives us two solutions for c:\newlineLower current: c=3−3=0c = 3 - 3 = 0 amperes\newlineHigher current: c=3+3=6c = 3 + 3 = 6 amperes

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