When a resistor is connected by itself to an ac generator, the average power del

Question

When a resistor is connected by itself to an ac generator, the average power delivered to the resistor is 0.840 W. When a capacitor is added in series with the resistor, the power delivered is 0.372 W. When an inductor is added in series with the resistor (without the capacitor), the power delivered is 0.165 W. Determine the power dissipated when both the capacitor and the inductor are added in series with the resistor. Note: The ac current and voltage are rms values and power is an average value unless indicated otherwise.

Explanation / Answer

This seems to me to be an excessively complicated problem, heavy on arithmetic, light on electrical principles, therefore not really worth struggling with, but I will show you how a solution can be reached.

First some general points; the usual equations for calculating power from V, I and R in ac circuits yield average power if V and/or I are expressed as rms values; Xl = reactance of the inductor, Xc = reactance of the cap, R = resistance of resistor, V = supply voltage, I = supply current.

1) Resistance alone: V^2 = 0.84*R ... (1)

2) R & C: I = V/sqrt(Xc^2 + R^2)

P = I^2*R = V^2*R/(Xc^2 + R^2) = 0.372

Substitute from (1) to get 0.84*R^2 = 0.372(Xc^2 + R^2)

rearrange this to get an equation (2) with Xc^2 as the subject expressed in terms of R^2

3) repeat step (2) to get an equation (3) with Xl^2 as the subject expressed in terms of R^2

4) With R, C & L in circuit:

P = I^2*R = V^2*R/|(Xl^2 - Xc^2 + R^2)| = 0.402

substitute the previously determined expressions for Xc^2 & Xl^2 to get an equation for P solely in terms of V & R

Finally solve this for P as a simultaneous pair with (1)

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.