Question

An L-C-R series circuit with 100Omega resisance is connected to an AC source of 200V and angular frequency 300rad//s. When only the capacitance is removed, the current lags behind the voltage by 60^@. When only the inductance is removed the current leads the voltage by 60^@. Calculate the current and the power dissipated in the L-C-R circuit

Answer 1

Given :

Resistance = 1000Ω

when capacitance is removed current lags behind by 60°

When inductance is removed current leads by 60°

To Find :

Current & power dissipated in LCR circuit

Solution :

•let phase difference is ¢

•when resistor and inductor are connected :

¢ = π/3

also , ¢ = tan^-1( XL-Xc)/R

π/3 = tan^-1( XL )/R (as Xc = 0)

XL = √3R ________(1)

•when resistor and capacitor are connected :

¢ = -π/3

also , ¢ = tan^-1(-Xc)/R

-π/3 = tan^-1( -Xc )/R (as XL = 0)

Xc = √3R ________(2)

•When L, C and R are connected in series

¢ = tan^-1 ( √3R-√3R)/R

¢ = 0

Also Z = √[(XL-Xc)²+R²]

Z = √[(√3R-√3R)²+R²]

Z = R

Z = 1000Ω

•Now P avg = V²rms cos¢/Z

Pavg =( V²rms cos0°)/R

Pavg = V²rms/R

Pavg = 200×200/1000

Pavg = 40 W

•Hence average power will be 40 W

• Current is given by :

I = V/Z = 200/1000 = 0.2 A

• Current in LCR circuit is given by 0.2 A