How Does AC Power Flow Through a Resistor, Inductor, and Capacitor in a Circuit?

[TMO]

Homework Statement

A circuit is constructed with an AC generator, a resistor, capacitor, and inductor as shown. The generator voltage varies in time ε=V_b-V_a=ε_m cos(ωt), where εm = 122.5 V and ω = 10.35 x 10⁶ radians/second. The resistance of the circuit is R= 132.1 Ω. The inductance is L= 10.3 μH. X_C = 533.060668709571.

https://www.smartphysics.com/Content/Media/UserData/8bc467c3-3c16-0dc8-ccd8-e71529e22ba1/noah_AC1.png

Homework Equations

$$ X_L = \omega L $$
$$ \tan(\phi) = \frac{X_L - X_C}{R} $$
$$ P(t) = I(t)^2R $$

The Attempt at a Solution

Because $$ I(t) = \frac{V(t)\cos(\phi}{R}) $$, therefore $$ P(t) = I(t)^2R = \frac{\epsilon_m^2\cos(\omega t)^2\cos(\phi)^2}{R}. $$ Using the second relevant equation to get phi, I should obtain the power at P(0) simply by plugging in values. However, the correct value isn't generated.

 

[ehild]

When we speak about power in an AC circuit, we mean the time average of the instantaneous power. When you integrate I²R with respect to time for a whole period and divide it by the time period T, you get that P_av=RI₀^2/2. (Io/√2 is the rms (root-mean-square) value of the AC current.)


ehild

 

[TMO]

Perhaps I haven't clarified. I need to know the magnitude of the power dissipated in the resistor at t=0.

 

[ehild]

φ=arctan((X_L-X_C)/R) is the phase of the voltage with respect to the current in the whole circuit. The voltage leads the current by the angle φ or the current lags behind the voltage by φ.
If the voltage is ε(t)=ε_m cos(ωt), the current is I₀cos((ωt-φ).
The amplitude of the voltage and current are related through the magnitude of the impedance _{Z=\sqrt{(X_L-X_C)^2+R^2}}.

ε_m=ZI₀. The instantaneous power is P(t)=I²R.

ehild

 

[gneill]

If, by t=0 one means that the power is switched on at time zero, so that the implied stimulus for the circuit is E_mcos(ωt)u(t), then you've got a situation with a transient response since the initially inert circuit will be "hit" with a step of E_m volts at that time.

Fortunately, the fact that it's a series circuit with an inductor helps you out in finding the initial current...