Question

A resistor R and an inductor L are connected in series to a source of voltage
V = V sin omegat. The voltage is found to lead current in phase by π/4. If the
inductor is replaced by a capacitor C, the voltage lags behind current in
phase by π/4. When L, C and R are connected in series with the same source,
Find the :
(1) average power dissipated and
(ii) instantaneous current in the circuit.​

Answer 1

Answer:

Explanation:

Power dicipated

=Vrms×Irms×cos(pi/8)

Instantaneous current equation

=i=i°cos(wt-pi/8)

Answer 2

Average power will be V²rms/R

Instantaneous current equation will be I. sinwt

•let phase difference is ¢

•when resistor and inductor are connected :

¢ = π/4

also , ¢ = tan^-1( XL-Xc)/R

π/4 = tan^-1( XL )/R (as Xc = 0)

XL = R ________(1)

•when resistor and capacitor are connected :

¢ = -π/4

also , ¢ = tan^-1(-Xc)/R

-π/4 = tan^-1( XL )/R (as Xc = 0)

Xc = R ________(2)

•When L, C and R are connected in series

¢ = tan^-1 ( R-R)/R

¢ = 0

Also Z = √[(XL-Xc)²+R²]

Z = √[(R-R)²+R²]

Z = R

•Now P avg = V²rms cos¢/Z

Pavg =( V²rms cos0°)/R

Pavg = V²rms/R

•Hence average power will be V²rms/R

•For instantaneous curtent equation

I(t) = V(t)/Z

I(t) = ( V. sinwt)/R

I(t) = I. sinwt