If ig(t) 0.5 cos 2000t A, find the average power absorbed by each element ((a) 130-Ω resistor, (b) 40-Ω resistor, (c) source, (d) inductor, (e) capacitor) in the circuit in the figure below.
Let's just go through it the straight way. You know, from your source, that: ? ? ? ? ω 2000 ? ? ? ? Ip ?A, ∴ Irms Ip ? (√2) You can compute the reactance of your capacitor (C12.5μF) and inductor (L60mH) as: ? ? ? ? XC 1 ? (ω?C) 40?, ∴ Z?C -40j ? ? ? ? XL ω?L 120?, ∴ Z?L 120j Let's say that: ? ? ? ? R? 130? ? ? ? ? R? 40? You have two parallel legs, each with a series pair of devices. The total impedance is in the capacitive leg is: ? ? ? ? Z? R? + Z?C 130 - 40j, ∴ |Z?| 10√185 In the inductive leg it is: ? ? ? ? Z? R? + Z?L 40 + 120j, ∴ |Z?| 40√10 So the total parallel impedance of these two legs is: ? ? ? ? Ztot Z??Z? ? (Z? + Z?) 200?(141 + 79j) ? 353 ? ? ? ? ∴ |Ztot| ≈ 91.57112 You know that: ? ? ? ? Vrms Irms ? |Ztot| ? ? ? ? Irms? Irms ? |Ztot| / |Z?| ? ? ? ? Irms? Irms ? |Ztot| / |Z?| ? ? ? ? P? Irms???R? (Irms ? |Ztot| / |Z?|)? ? R? 7??????? W ≈ 7.365 W ? ? ? ? P? Irms???R? (Irms ? |Ztot| / |Z?|)? ? R? 2??????? W ≈ 2.620 W ? ? ? ? Ptot P? + P? 9??????? W ≈ 9.986 W Real power isn't dissipated in ideal capacitors or inductors, so (a) ≈ 7.365 W, (b) ≈ 2.620 W, (c) ≈ 9.986 W, (d) 0 W, and (e) 0 W. Since the problem you show doesn't illustrate VAR units on it, I don't think (d) and (e) should use reactive power. ? ? ? ? P? Irms???R? (Irms ? |Ztot| / |Z?|)? ? R? 7??????? W ≈ 7.365 W ? ? ? ? P? Irms???R? (Irms ? |Ztot| / |Z?|)? ? R? 2??????? W ≈ 2.620 W The above was validated with Spice.
