Question

A sinusoidal voltage of peak value 283 v and frequency 50 hz is applied to a series lcr circuit in which r

Answer 1

Answer:

(a) impedance of circuit is 5 ohm

(b) Phase angle of circuit is 53.1°

(c) Power dissipated in circuit is 4800 watt

(D) power factor of circuit is 0.6

Explanation:

Given as :

For A series LCR circuit

Peak voltage = 283 volt

Frequency = f = 50 Hz

According to question

(a)

Impedance = Z = $\sqrt{R^{2} + (X_l - X_c)^{2} }$

Or, Z = $\sqrt{R^{2} + (X_l - X_c)^{2} }$

$X_l$ = Lω

Or, $X_l$ = 2 × 3.14 × 50 × 25.8 × $10^{-3}$

Or, $X_l$ = 8 ohm

And

$X_c$  = $\dfrac{1}{wc}$

Or, $X_c$ = $\dfrac{1}{2\times 3.14\times 50\times 796\times 10^{-6}}$

Or, $X_c$ = 5 ohm

∴  Z = $\sqrt{3^{2} + (8 - 4)^{2} }$

i.e impedance = 5 ohm

(b) Phase difference

∅ = $tan^{-}(\dfrac{X_l-X_c}{R})$

Or, Ф  = $tan^{-}(\dfrac{8-4}{3})$

∴ phase angle   = 53.1°

(c)  power dissipated

Power = I² R

$V_r_m_s$  = $\dfrac{V_0}{\sqrt{2} }$  = $\dfrac{283}{\sqrt{2} }$  volt

Or, P =( $\dfrac{283}{R\sqrt{2} }$  )² × R

Or, power = 4800 watt

(d) power factor = cosФ

Or, cosФ = cos 53.1°

∴ Power factor = 0.6

Hence,

(a) impedance of circuit is 5 ohm

(b) Phase angle of circuit is 53.1°

(c) Power dissipated in circuit is 4800 watt

(D) power factor of circuit is 0.6  Answer