The average dissipated power in a resistor in a ac circuit is:
[tex]P=I_{rms}^2 R[/tex]
where R is the resistance, and [tex]I_{rms}[/tex] is the root mean square current, defined as
[tex]I_{rms} = \frac{I_0}{\sqrt{2}} [/tex]
where [tex]I_0[/tex] is the peak value of the current. Substituting the second formula into the first one, we find
[tex]P=( \frac{I_0}{\sqrt{2} } )^2 R = \frac{1}{2} I_0^2 R [/tex]
and if we re-arrange this formula and use the data of the problem, we can find the value of the peak current I0:
[tex]I_0 = \sqrt{ \frac{2 P}{R} }= \sqrt{ \frac{2 \cdot 2.0 W}{47 \Omega} }=0.29 A [/tex]
