LCR Parallel Circuit
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The following quantities appear in the problem:
Impedanz \(Z\) / Zeit \(t\) / Amplitude \(\hat y\) / Winkelgeschwindigkeit / Kreisfrequenz \(\omega\) / Dämpfungskoeffizient \(\delta\) / Phasenverschiebung \(\varphi\) / Kapazität \(C\) / Blindwiderstand \(X\) / Selbstinduktivität \(L\) / elektrischer Widerstand \(R\) /
The following formulas must be used to solve the exercise:
\(X_C = \frac{1}{\omega C} \quad \) \(X_L = \omega L \quad \) \(\dfrac{1}{Z} = \sqrt{\dfrac{1}{R^2}+\left(\dfrac{1}{X_L}-\dfrac{1}{X_C}\right)^2} \quad \) \(Z = \sqrt{R^2+(X_L-X_C)^2} \quad \) \(\omega = \frac{1}{\sqrt{LC}} \quad \) \(\tan\varphi = R\left(\dfrac{1}{X_L}-\dfrac{1}{X_C}\right) \quad \) \(\tan\varphi = \dfrac{X_L-X_C}{R} \quad \) \(A_t = A_0 \cdot \text{e}^{-\delta t} \quad \)
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Exercise:
Show that the impedance and phase shift of an ac circuit with a resistor resistance R a capacitor capacitance C and a coil inductance L in parallel are given by * fracZ sqrtfracR^+leftomega C-fracomega Lright^ and * tanDeltaphi Rleftfracomega L-omega Cright H: Since the voltage is the same across all three elements in the parallel circuit you can start with the corresponding phasor in a phasor diagram and add the current phasors with their respective phaseshift.
Solution:
In the phasor diagram the current phasor for the resistor is in phase with the voltage while the current phasors for the capacitor and the coil have a phase shift of mp fracpi respectively see figure. center includegraphicswidth.mm#image_path:phasors-parallel-circuit# center The phasor for the total current I can be found by adding the phasors for the partial currents I_R I_C and I_L as vectors. The amplitude of the total current is I sqrtI_R^+leftI_C-I_Lright^ Using the relations I_R fracVR I_C fracVX_C I_L fracVX_L it follows for the impedance fracZ fracIV fracsqrtleftfracVRright^+leftfracVX_C-fracVX_Lright^V sqrtfracR^+leftfracX_C-fracX_Lright^ With the expressions for the reactances X_C and X_L we find fracZomega sqrtfracR^+leftomega C-fracomega Lright^ The phase shift is given by see figure tanDeltaphi fracI_L-I_CI_R fracfracVX_L-fracVX_CfracVR RleftfracX_L-fracX_Cright Rleftfracomega L-omega Cright Alternatively the expressions can be derived using the complex reactances tildeX_L and tildeX_C. For a parallel circuit the reciprocal value of the total complex impedance corresponds to the of the reciprocal values of the partial values: fractildeZ fracR+fractildeX_L+fractildeX_C fracR+fracjomega L+jomega CfracR-fracjomega L+jomega C fracR+jleftomega C-fracomega Lright The complex impedance is tildeZ fracfracR+jleftomega C-fracomega Lright fracfracR-jleftomega C-fracomega LrightfracR^+leftomega C-fracomega Lright^ The real impedance is given by the modulus of the complex impedance: Z sqrtRetildeZ^+ImtildeZ^fracsqrtfracR^+leftomega C-fracomega Lright^fracR^+leftomega C-fracomega Lright^ fracsqrtfracR^+leftomega C-fracomega Lright^ The phase shift corresponds to the argument: tanDeltaphi fracImtildeZRetildeZ fracfracomega L-omega CfracR Rleftfracomega L-omega Cright
Show that the impedance and phase shift of an ac circuit with a resistor resistance R a capacitor capacitance C and a coil inductance L in parallel are given by * fracZ sqrtfracR^+leftomega C-fracomega Lright^ and * tanDeltaphi Rleftfracomega L-omega Cright H: Since the voltage is the same across all three elements in the parallel circuit you can start with the corresponding phasor in a phasor diagram and add the current phasors with their respective phaseshift.
Solution:
In the phasor diagram the current phasor for the resistor is in phase with the voltage while the current phasors for the capacitor and the coil have a phase shift of mp fracpi respectively see figure. center includegraphicswidth.mm#image_path:phasors-parallel-circuit# center The phasor for the total current I can be found by adding the phasors for the partial currents I_R I_C and I_L as vectors. The amplitude of the total current is I sqrtI_R^+leftI_C-I_Lright^ Using the relations I_R fracVR I_C fracVX_C I_L fracVX_L it follows for the impedance fracZ fracIV fracsqrtleftfracVRright^+leftfracVX_C-fracVX_Lright^V sqrtfracR^+leftfracX_C-fracX_Lright^ With the expressions for the reactances X_C and X_L we find fracZomega sqrtfracR^+leftomega C-fracomega Lright^ The phase shift is given by see figure tanDeltaphi fracI_L-I_CI_R fracfracVX_L-fracVX_CfracVR RleftfracX_L-fracX_Cright Rleftfracomega L-omega Cright Alternatively the expressions can be derived using the complex reactances tildeX_L and tildeX_C. For a parallel circuit the reciprocal value of the total complex impedance corresponds to the of the reciprocal values of the partial values: fractildeZ fracR+fractildeX_L+fractildeX_C fracR+fracjomega L+jomega CfracR-fracjomega L+jomega C fracR+jleftomega C-fracomega Lright The complex impedance is tildeZ fracfracR+jleftomega C-fracomega Lright fracfracR-jleftomega C-fracomega LrightfracR^+leftomega C-fracomega Lright^ The real impedance is given by the modulus of the complex impedance: Z sqrtRetildeZ^+ImtildeZ^fracsqrtfracR^+leftomega C-fracomega Lright^fracR^+leftomega C-fracomega Lright^ fracsqrtfracR^+leftomega C-fracomega Lright^ The phase shift corresponds to the argument: tanDeltaphi fracImtildeZRetildeZ fracfracomega L-omega CfracR Rleftfracomega L-omega Cright