# AC Power Thru a Resistor

1. Nov 25, 2012

### TMO

1. The problem statement, all variables and given/known data

A circuit is constructed with an AC generator, a resistor, capacitor, and inductor as shown. The generator voltage varies in time ε=Vb-Vam 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. XC = 533.060668709571.

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

2. Relevant equations

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

3. 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.

Last edited by a moderator: May 6, 2017
2. Nov 25, 2012

### 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 Pav=RI0^2/2. (Io/√2 is the rms (root-mean-square) value of the AC current.)

ehild

3. Nov 26, 2012

### TMO

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

4. Nov 26, 2012

### ehild

φ=arctan((XL-XC)/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)=εm cos(ωt), the current is I0cos((ωt-φ).
The amplitude of the voltage and current are related through the magnitude of the impedance $Z=\sqrt{(X_L-X_C)^2+R^2}$.

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

ehild

5. Nov 26, 2012

### Staff: Mentor

If, by t=0 one means that the power is switched on at time zero, so that the implied stimulus for the circuit is Emcos(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 Em volts at that time.

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