Question

An LCR series circuit with 100 resistance is connected to an AC source of 200 V and angular frequency 300 radians per second. When only the capacitance is removed, the current leads the voltage by 60 . Then the current and power dissipated in LCR circuit are respectively :

Answer 1

Given LCR circuit

Resistance $R=100r$

rms voltage v$=200$v

Angular frequency$=300$ radiant/sec

Current lags behind the voltage by $6{-}^o$

$\therefore \tan 6=\frac{\omega L}{R}$

$\tan {60}^o=\frac{\frac{1}{\omega c}}{R}$

$\omega L=\frac{1}{\omega C}$

Impedance of circuit

$z=\sqrt{R^2+{(\omega L-\frac{1}{\omega C})}^2}=R$

Current in circuit:

$z_0=\frac{v_o}{z}=\frac{v_o}{R}=\frac{200}{100}=2A=I$

Average power $P=\frac{1}{2}{v}_o{I}_o\cos \theta$

$\tan \theta =\frac{\omega L-\frac{1}{\omega C}}{R}=0(\cos \theta =1)$.

$\therefore P=\frac{1}{2}\times 200\times 2\times 1=200$.