II Review I Constants Consider the power dissipated in a series R-L-C circuit with R= 3702, L = 120mH, C = 0.300µF, V = 50V, and w = 12500rad/s. The current and the voltages in this circuit are not in phase with each other. Using the values given, the phase angle o was found to be 73 °, and the current amplitude I was found to be 3.9x10-2 A. Calculate the power factor and the average power to the entire circuit and to each circuit element. SOLUTION SET UP AND SOLVE The power factor is cos o = cos 73° = 0.29. The average power to the circuit is P=}VI cos o =(50V)(3.9 × 10-²A)(0.29) = 0.28 W REFLECT AIl of this power is dissipated in the resistor; the average power to a pure inductor or pure capacitor is always zero. Part A - Practice Problem: If the inductance L in this circuit could be changed, what value of L would give a power factor of unity? Express your answer with the appropriate units. ? Value Units Submit Request Answer

II Review I Constants Consider the power dissipated in a series R-L-C circuit with R= 3702, L = 120mH, C = 0.300µF, V = 50V, and w = 12500rad/s. The current and the voltages in this circuit are not in phase with each other. Using the values given, the phase angle o was found to be 73 °, and the current amplitude I was found to be 3.9x10-2 A. Calculate the power factor and the average power to the entire circuit and to each circuit element. SOLUTION SET UP AND SOLVE The power factor is cos o = cos 73° = 0.29. The average power to the circuit is P=}VI cos o =(50V)(3.9 × 10-²A)(0.29) = 0.28 W REFLECT AIl of this power is dissipated in the resistor; the average power to a pure inductor or pure capacitor is always zero. Part A - Practice Problem: If the inductance L in this circuit could be changed, what value of L would give a power factor of unity? Express your answer with the appropriate units. ? Value Units Submit Request Answer

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Review I Constants
Consider the power dissipated in a series R-L-C circuit with
R= 370N, L = 120mH, C = 0.300µF, V = 50V, and
w = 12500rad/s. The current and the voltages in this circuit are
not in phase with each other. Using the values given, the phase
angle o was found to be 73° , and the current amplitude I was
found to be 3.9x10-2 A . Calculate the power factor and the
average power to the entire circuit and to each circuit element.
%3D
SOLUTION
SET UP AND SOLVE The power factor is cos o = cos 73° = 0.29. The average power to the circuit is
P=¿VI cos o =:(50V)(3.9 × 10-2A)(0.29) = 0.28 W
REFLECT All of this power is dissipated in the resistor; the average power to a pure inductor or pure capacitor is always zero.
Part A - Practice Problem:
If the inductance L in this circuit could be changed, what value of L would give a power factor of unity?
Express your answer with the appropriate units.
HA
?
Value
Units
Submit
Request Answer

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