91Ó°ÊÓ

The rms voltage of the AC source is:\(\Delta {V_{rms }} = 120{\rm{\;V}}\)

The rms current in the circuit is: \({I_{rms}} = 10.0\;\;{\rm{A}}\)

The average power generated by an AC source to a resistive load is:

\(\begin{array}{c}{{\bf{P}}_{{\bf{avg}}}}{\bf{ = }}{{\bf{I}}_{{\bf{rms}}}}{\bf{\Delta }}{{\bf{V}}_{{\bf{rms}}}}\\{\bf{ = }}\frac{{\bf{1}}}{{\bf{2}}}{{\bf{I}}_{{\bf{max}}}}{\bf{\Delta }}{{\bf{V}}_{{\bf{max}}}}\end{array}\)

Where\({{\rm{I}}_{{\rm{rms}}}}\)is the rms current,\({\rm{\Delta }}{{\rm{V}}_{{\rm{rms }}}}\)is the rms voltage,\({{\rm{I}}_{{\rm{max}}}}\)is the peak current, and\({\rm{\Delta }}{{\rm{V}}_{{\rm{max}}}}\)is the peak voltage.

The peak power delivered consumption of a circuit element with an AC voltage source is:

\({{\bf{P}}_{{\bf{max}}}}{\bf{ = }}{{\bf{I}}_{{\bf{max}}}}{\bf{\Delta }}{{\bf{V}}_{{\bf{max}}}}\)

The average power consumption of the microwave oven is found from equation (1):

\({P_{avg}} = {I_{rms}}\Delta {V_{rms}}\)

Entering the values for\({I_{rms}}\) and\(\Delta {V_{rms}}\)it gives:

\(\begin{array}{c}{P_{avg}} = (10.0\;{\rm{A}})(120\;{\rm{V}})\\ = 1200\;{\rm{W}}\end{array}\)

The maximum consumption of power done by the microwave oven is found using equation (2):

\({P_{\max }} = {I_{\max }}\Delta {V_{\max }}\)

Multiply the right-hand side by 2 and divide it by 2:

\({P_{\max }} = 2\left( {\frac{1}{2}{I_{\max }}\Delta {V_{\max }}} \right)\)

For the given values, Equation (1) becomes:

\(\begin{array}{c}{P_{\max }} = 2{P_{avg }}\\ = 2(1200\;{\rm{W}})\\ = 2400\;{\rm{W}}\\ = 2.40\;{\rm{kW}}\end{array}\)