I′=2∠0∘×−j0.410−j1−j.01=79.22∠−82.03∘mA
I′=79.22cos(5t−82.03∘)mA
I′′=(5∠0∘)×(−j1.6667)10−j0.6667−j1.6667
=811.5∠−76.86∘mA
I=I′+I′′=[79.22cos(5t−82.03∘)+8115.cos(3t−76.85∘)]mA
Rms value of this current, I2rms=(79.22×10−3)22+(811.5×10−3)22=0.332
Power dissipated by 10Ω resistor,
P=I2rms×R
=0.332×10=3.32W