Question 23366
x1(t)=Acos(ωt)=Acos(2πλCt),I=A2.x2(t)=3Acos(ωt+δ)=3Acos(2πλCt+δ),I=9A2.x(t)=x1(t)+x2(t)=Acos(2πλCt)+3A[cos2πλCt⋅cosδ−sin2πλCt⋅sinδ]
For δ=0:x(t)=Acos(2πλCt)+3A[cos2πλCt]=4Acos2πλCt,I=16A2.
For δ=π:x(t)=Acos(2πλCt)−3A[cos2πλCt]=−2Acos2πλCt,I=4A2.