Question

a capacitor of unknown capacitance, a resistor of 100 ohm and an inductor of self inductance L=(4/pi^2) henry are connected in series to an ac source of 200V and 50Hz. Calculate the value of the capacitance and impedance of the circuit when the current is in phase with the voltage. Calculate the power dissipated in the circuit

Answer 1

Current in phase with voltage means angle between emf and current is 0 . This is resonance condition 
in resonance, inductive reactance = capacitive reactance.
=> X(L) = x(c)
    wL = 1/(wc)    ( here w is omega)
   so C = 1/(w^2 L )

Sustitute value of L ( given ) and w= 2(pi) f = 2pi ×50

you get c= 1/(30000)  farad

impedance of resonating circuit = resistance  = 100 ohm

power = vI = V^2/R = (200)^2 / 100 = 400 watt
hope this helps you..

Answer 2

Hey !!

R = 100 Ω , L = 4/π² H , C , Vrms = 200 V , f = 50 Hz , C = ? , Z = ?

Ф = 0° , power = ?

When current and voltage are in phase

     Lw = 1/CW  => (4/π²) (2 × π × 50) = 1/C (2π × 50)

=> 8 × 50 × 100 = 1/C

=> 4 × 10⁴ = 1/C

=> C = 10⁻⁴ / 4

              ∴ C = 2.5 × 10⁻⁵ F = 25 μF

Z = √R² + (LW - 1/CW)² = √R² = R                   (∵ Lw = 1/Cw)

     ∴ Z = 100 Ω

irms = 200 / 100 = 2A

∴ Power = 200 × 2 × 100

        = 400 W

Good luck !!