PhysicsMedium94×since 2002Q5870In an a.c.a.c.a.c. circuit the voltage applied is E=E0 sin ωt.E = {E_0}\,\sin \,\omega t.E=E0sinωt. The resulting current in the circuit is I=I0sin(ωt−π2).I = {I_0}\sin \left( {\omega t - {\pi \over 2}} \right).I=I0sin(ωt−2π). The power consumption in the circuit is given byAP=2E0I0P = \sqrt 2 {E_0}{I_0}P=2E0I0BP=E0I02P = {{{E_0}{I_0}} \over {\sqrt 2 }}P=2E0I0CP=zeroP=zeroP=zeroDP=E0I02P = {{{E_0}{I_0}} \over 2}P=2E0I0Check answerSkip