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Express the sinusoidal voltage V(t)=10cos(40t)V(t)=10\cos(40t)V(t)=10cos(40t) as a complex phasor V=A∠ϕV= A\angle\phiV=A∠ϕ.
Convert the voltage V(t)=5cos(40t+90∘)V(t)=5\cos(40t+90^\circ)V(t)=5cos(40t+90∘) into its complex exponential form V(t)=ℜ{V ej40t}V(t)=\Re\{V\,e^{j40t}\}V(t)=ℜ{Vej40t} and state the phasor VVV.
Three balanced three-phase voltages are Va=100cos(60t),Vb=100cos(60t−120∘),Vc=100cos(60t+120∘).V_a=100\cos(60t),\quad V_b=100\cos(60t-120^\circ),\quad V_c=100\cos(60t+120^\circ).Va=100cos(60t),Vb=100cos(60t−120∘),Vc=100cos(60t+120∘). Show that their sum is zero by phasor addition.
Given two sinusoidal voltages V1(t)=10cos(40t),V2(t)=20cos(40t+30∘),V_1(t)=10\cos(40t),\quad V_2(t)=20\cos(40t+30^\circ),V1(t)=10cos(40t),V2(t)=20cos(40t+30∘), find the resulting voltage in the form V(t)=Acos(40t+B).V(t)=A\cos(40t+B).V(t)=Acos(40t+B).
Compute the sum V(t)=10cos(40t+350∘)+10cos(40t−10∘)V(t)=10\cos(40t+350^\circ)+10\cos(40t-10^\circ)V(t)=10cos(40t+350∘)+10cos(40t−10∘) and express it as Acos(40t+B)A\cos(40t+B)Acos(40t+B).
Two voltages are given by V1=20cos(40t+170∘),V2=20cos(40t−170∘).V_1=20\cos(40t+170^\circ),\quad V_2=20\cos(40t-170^\circ).V1=20cos(40t+170∘),V2=20cos(40t−170∘). Find the resultant voltage V(t)=V1+V2V(t)=V_1+V_2V(t)=V1+V2 in the form Acos(40t+B)A\cos(40t+B)Acos(40t+B).
Derive an expression for the sum V(t)=Vmcos(ωt+α)+Vmcos(ωt−α)V(t)=V_m\cos(\omega t+\alpha)+V_m\cos(\omega t-\alpha)V(t)=Vmcos(ωt+α)+Vmcos(ωt−α) in the form V(t)=Acos(ωt+ϕ)V(t)=A\cos(\omega t+\phi)V(t)=Acos(ωt+ϕ) and find AAA in terms of VmV_mVm and α\alphaα.
Given V1(t)=15cos(50t−20∘)andV2(t)=25cos(50t+40∘),V_1(t)=15\cos(50t-20^\circ)\quad\text{and}\quad V_2(t)=25\cos(50t+40^\circ),V1(t)=15cos(50t−20∘)andV2(t)=25cos(50t+40∘), determine the amplitude AAA and phase BBB of V(t)=V1(t)+V2(t)=Acos(50t+B).V(t)=V_1(t)+V_2(t)=A\cos(50t+B).V(t)=V1(t)+V2(t)=Acos(50t+B).
Sum three sinusoidal voltages of the same frequency: V1=10cos(100t),V2=10cos(100t+120∘),V3=10cos(100t+240∘).V_1=10\cos(100t),\quad V_2=10\cos(100t+120^\circ),\quad V_3=10\cos(100t+240^\circ).V1=10cos(100t),V2=10cos(100t+120∘),V3=10cos(100t+240∘). Express the total voltage as V(t)=Acos(100t+B)V(t)=A\cos(100t+B)V(t)=Acos(100t+B).
A sinusoid has phasor representation V=12+j5V=12+j5V=12+j5. Convert this to time-domain form v(t)=Acos(40t+ϕ),v(t)=A\cos(40t+\phi),v(t)=Acos(40t+ϕ), giving AAA and ϕ\phiϕ in degrees.
Given phasors V1=5∠30∘V_1=5\angle30^\circV1=5∠30∘, V2=12∠(−45∘)V_2=12\angle(-45^\circ)V2=12∠(−45∘), and V3=8∠60∘V_3=8\angle60^\circV3=8∠60∘ at ω=40 \omega=40\,ω=40rad/s, find the time-domain sum v(t)=ℜ{(V1+V2+V3)ej40t}v(t)=\Re\{(V_1+V_2+V_3)e^{j40t}\}v(t)=ℜ{(V1+V2+V3)ej40t} in the form Acos(40t+B)A\cos(40t+B)Acos(40t+B).
Express the difference V(t)=30cos(40t+45∘)−50cos(40t−15∘)V(t)=30\cos(40t+45^\circ)-50\cos(40t-15^\circ)V(t)=30cos(40t+45∘)−50cos(40t−15∘) in the form Acos(40t+B)A\cos(40t+B)Acos(40t+B).
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