JEE Class main Answered

Resistance in the circuit , R = 30 Ω .

Inductive reactance in the circuit , Lω = 40 Ω

Impedance z in the circuit , z = [ 30² + 40² ]^1/2 = 50 Ω .

Voltage of alternating source , E = ( 200 V ) sin ( 50π t )

Maximum magnitude of current through resistor, i_m  = ( Maximum voltage ) / Impedance

Maximum magnitude of current through resistor, i_m =   (200  V )/ ( 50 Ω ) = 4 A

Instantaneous current through resistor , i = i_m sin ( 50π t + φ )

where φ is phase difference between voltage and current

φ = tan^-1( Lω / R ) = tan^-1 ( 40/30) = 53°

Power dissipated across resistor at time t is

P_R ( t ) = i² R = i_m² R  sin² ( 50π t + φ )

Power dissipated across resistor at time t= 1 s is

P_R ( 1 ) = ( 16 × 30 W )  sin² ( 50π  + φ ) = ( 480 W )  × sin² ( 53° ) = 306 W