RL parallel circuit
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Exercise:
Calculate the impedance and the phase shift for an ac circuit with a resistor resistance RO and a coil inductive reactance XL in parallel.
Solution:
Since the resistor and the coil are connected in parallel the voltage V across them is the same; we use it as the reference phasor. The current through the resistor I_R V/R is in phase with the voltage while the current through the coil I_L V/X_L lags the voltage by degree opposite to a capacitor. The total current I is the phasor of I_R and I_L. center tikzpicturestealth scale. draw- - -- noderight mathrmRe; draw- - -- nodeabove mathrmIm; draw- red thick -- nodemidway above V; draw- blue thick -- nodemidway below I_R; draw- blue thick -- - nodemidway right I_L; draw- blue ultra thick -- - nodemidway below left I; draw arc :-.:; node at - varphi; tikzpicture center Dividing every phasor by the real in-phase voltage V turns the current triangle o the reciprocal impedance triangle since I/V /Z I_R/V /R and I_L/V /X_L. From this right triangle we obtain the magnitude of the total current and from it the total impedance as well as the phase angle varphi between the voltage V and the current I: I sqrtI_R^+I_L^ VsqrtfracR^+fracX_L^ Z fracVI fracVVsqrtdfracR^+dfracX_L^ fracsqrtdfracR^+dfracX_L^ fracR X_LsqrtR^+X_L^ fracRtimesXLsqrtR^+XL^ Z approx resultZP varphi arctanleftfracI_LI_Rright arctanleftfracRX_Lright arctanleftfracRXLright ph approx resultphP medskip Alternatively we can assign a complex quantity to each element. The resistor's contribution is purely real R while the coil's complex reactance is tilde X_L iX_L since the current through an inductor lags the voltage by degree. Because the two branches are in parallel it is the complex admittances that add: tilde Y fracR + fractilde X_L fracR - fraciX_L The complex impedance is the reciprocal of tilde Y: tilde Z fractilde Y fracdfracR - idfracX_L Its magnitude and argument reproduce exactly the impedance and phase shift found above: Z |tilde Z| fracsqrtdfracR^+dfracX_L^ fracR X_LsqrtR^+X_L^ Z approx resultZP varphi argtilde Z -argtilde Y arctanleftfracRX_Lright ph approx resultphP confirming the result obtained with the phasor diagram.
Calculate the impedance and the phase shift for an ac circuit with a resistor resistance RO and a coil inductive reactance XL in parallel.
Solution:
Since the resistor and the coil are connected in parallel the voltage V across them is the same; we use it as the reference phasor. The current through the resistor I_R V/R is in phase with the voltage while the current through the coil I_L V/X_L lags the voltage by degree opposite to a capacitor. The total current I is the phasor of I_R and I_L. center tikzpicturestealth scale. draw- - -- noderight mathrmRe; draw- - -- nodeabove mathrmIm; draw- red thick -- nodemidway above V; draw- blue thick -- nodemidway below I_R; draw- blue thick -- - nodemidway right I_L; draw- blue ultra thick -- - nodemidway below left I; draw arc :-.:; node at - varphi; tikzpicture center Dividing every phasor by the real in-phase voltage V turns the current triangle o the reciprocal impedance triangle since I/V /Z I_R/V /R and I_L/V /X_L. From this right triangle we obtain the magnitude of the total current and from it the total impedance as well as the phase angle varphi between the voltage V and the current I: I sqrtI_R^+I_L^ VsqrtfracR^+fracX_L^ Z fracVI fracVVsqrtdfracR^+dfracX_L^ fracsqrtdfracR^+dfracX_L^ fracR X_LsqrtR^+X_L^ fracRtimesXLsqrtR^+XL^ Z approx resultZP varphi arctanleftfracI_LI_Rright arctanleftfracRX_Lright arctanleftfracRXLright ph approx resultphP medskip Alternatively we can assign a complex quantity to each element. The resistor's contribution is purely real R while the coil's complex reactance is tilde X_L iX_L since the current through an inductor lags the voltage by degree. Because the two branches are in parallel it is the complex admittances that add: tilde Y fracR + fractilde X_L fracR - fraciX_L The complex impedance is the reciprocal of tilde Y: tilde Z fractilde Y fracdfracR - idfracX_L Its magnitude and argument reproduce exactly the impedance and phase shift found above: Z |tilde Z| fracsqrtdfracR^+dfracX_L^ fracR X_LsqrtR^+X_L^ Z approx resultZP varphi argtilde Z -argtilde Y arctanleftfracRX_Lright ph approx resultphP confirming the result obtained with the phasor diagram.