Question

In an alternating current circuit a voltage source V = 100 sin (100 pit), a resistance of 10 ohm and an inductance of 1/(10 pi) henry, all are connected in series. Find out the power dissipated in the circuit: (1) 300 watt (2) 400 watt (3) 500 watt (4) 250 watt'

Answer 1

Series R L circuit.  with AC  voltage source.

V = 100 Sin (100π t)
     ω = 100 π  rad/sec        => f = ω/2π = 50 Hz

   R = 10 Ω  ,          L = 0.1/π Henry  
  X_L = inductive reactance = ω L = 100π 0.1/π = 10 Ω

   Z = impedance in the circuit =  √[ R² + X_L² ] = √[10²+10²] = 10√2 Ω

  V_rms = V_max / √2  =  100/√2  volts 
  I_rms = V_rms / Z  = 100/(√2 10√2) = 5 Amperes

  Power discharged by the voltage source =  V_rms  I_rms =  500/√2  VA
             (all this power is not lost.. energy in Inductor is not lost)

  Real power discharged (dissipated) by the resistor as heat = I_rms² R
                   = 5² * 10 = 250 Watts

Alternately:

  TanФ = phase difference between currents in L and R:  X_L / R  = 1
       Ф = π/4
  Cos Ф = power factor =   Cos π/4 = 1/√2

   Real power = Reactive power * power factor = 500/√2 VA * cosФ = 250 Watts
 
Power is stored and then released alternately in the inductor... Dissipation takes place only in  the resistor in all RLC circuits.