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Exercise:
An inductor is connected in series to a capacitor and a resistor in parallel. Derive the formal expressions impedance and phase shift. Visualize how the impedance and the phase shift dep on the angular frequency. center includegraphicswidthcm#image_path:l-rc-# center

Solution:
Using the complex reactances for the inductor and the capacitor the complex impedance can be calculated as follows: tildeZ tilde X_L + leftfracR + fractilde X_C right^- j omega L + leftfracR + j omega C right^- j omega L + leftfrac+j omega C RR right^- j omega L + fracR + j omega R C j omega L + fracR - j omega R C+j omega R C-j omega R C j omega L + fracR - j omega R C + omega^ R^ C^ fracR+omega^ R^ C^ + j leftomega L - fracomega R C+omega^ R^ C^ right frac+omega^ R^ C^ leftR + j leftomega L left+omega^ C^ R^ right-omega C R^ right right The impedance corresponds to the modulus: Z abstilde Z frac+omega^ R^ C^ sqrtR^ + leftomega L left+omega^ C^ R^ right-omega C R^ right^ frac+omega^ R^ C^ sqrtR^ + omega^ leftL left+omega^ C^ R^ right-C R^ right^ The phase shift is given by the argument: tanDeltaphi argtilde Z fracomega leftL left+omega^ C^ R^ right-C R^ rightR The graphs for Zomega and Deltaphiomega are displayed below. center includegraphicswidthcm#image_path:l-rc-impedanc# center center includegraphicswidthcm#image_path:l-rc-phasshift-# center
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Exercise:
An inductor is connected in series to a capacitor and a resistor in parallel. Derive the formal expressions impedance and phase shift. Visualize how the impedance and the phase shift dep on the angular frequency. center includegraphicswidthcm#image_path:l-rc-# center

Solution:
Using the complex reactances for the inductor and the capacitor the complex impedance can be calculated as follows: tildeZ tilde X_L + leftfracR + fractilde X_C right^- j omega L + leftfracR + j omega C right^- j omega L + leftfrac+j omega C RR right^- j omega L + fracR + j omega R C j omega L + fracR - j omega R C+j omega R C-j omega R C j omega L + fracR - j omega R C + omega^ R^ C^ fracR+omega^ R^ C^ + j leftomega L - fracomega R C+omega^ R^ C^ right frac+omega^ R^ C^ leftR + j leftomega L left+omega^ C^ R^ right-omega C R^ right right The impedance corresponds to the modulus: Z abstilde Z frac+omega^ R^ C^ sqrtR^ + leftomega L left+omega^ C^ R^ right-omega C R^ right^ frac+omega^ R^ C^ sqrtR^ + omega^ leftL left+omega^ C^ R^ right-C R^ right^ The phase shift is given by the argument: tanDeltaphi argtilde Z fracomega leftL left+omega^ C^ R^ right-C R^ rightR The graphs for Zomega and Deltaphiomega are displayed below. center includegraphicswidthcm#image_path:l-rc-impedanc# center center includegraphicswidthcm#image_path:l-rc-phasshift-# center
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Attributes & Decorations
Branches
Electrodynamics
Tags
ac circuit, complex impedance, impedance, phaseshift
Content image
Difficulty
(3, default)
Points
0 (default)
Language
ENG (English)
Type
Calculative / Quantity
Creator by
Decoration