LCR Series 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 series are given by * Z sqrtR^+leftfracomega C-omega Lright^ and * tanDeltaphi fracomega L-dfracomega CR H: Since the current is the same through all three elements in the series circuit you can start with the corresponding phasor in a phasor diagram and add the voltage phasors with their respective phaseshift.
Solution:
In the phasor diagram the voltage phasor for the resistor is in phase with the current while the voltage phasors for the capacitor and the coil have a phase shift of pm fracpi respectively see figure. center includegraphicswidthmm#image_path:phasor-diagram-series-circuit# center The phasor for the total voltage V can be found by adding the phasors for the partial voltages V_R V_C and V_L as vectors. The amplitude of the total voltage is V sqrtV_R^+leftV_C-V_Lright^ Using the relations V_R R I V_C X_C I V_L X_L I it follows for the impedance Z fracVI fracsqrtleftR Iright^+leftX_C I-X_L Iright^I sqrtR^+leftX_C-X_Lright^ With the expressions for the reactances X_C and X_L we find Zomega sqrtR^+leftfracomega C-omega Lright^ The phase shift is given by see figure tanDeltaphi fracV_L-V_CV_R fracX_L I-X_C IR I fracX_L-X_CR fracomega L-fracomega CR Alternatively the expressions can be derived using the complex reactances tildeX_L and tildeX_C. For a series circuit the total complex impedance corresponds to the of the partial values: tildeZ R+tildeX_L+tildeX_C R+jomega L+fracjomega C R+jomega L-fracjomega C R+jleftomega L-fracomega Cright The real impedance is given by the modulus of the complex impedance: Z sqrtRetildeZ^+ImtildeZ^ sqrtR^+leftomega L-fracomega Cright^ The phase shift corresponds to the argument: tanDeltaphi fracImtildeZRetildeZ fracomega L-fracomega CR
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 series are given by * Z sqrtR^+leftfracomega C-omega Lright^ and * tanDeltaphi fracomega L-dfracomega CR H: Since the current is the same through all three elements in the series circuit you can start with the corresponding phasor in a phasor diagram and add the voltage phasors with their respective phaseshift.
Solution:
In the phasor diagram the voltage phasor for the resistor is in phase with the current while the voltage phasors for the capacitor and the coil have a phase shift of pm fracpi respectively see figure. center includegraphicswidthmm#image_path:phasor-diagram-series-circuit# center The phasor for the total voltage V can be found by adding the phasors for the partial voltages V_R V_C and V_L as vectors. The amplitude of the total voltage is V sqrtV_R^+leftV_C-V_Lright^ Using the relations V_R R I V_C X_C I V_L X_L I it follows for the impedance Z fracVI fracsqrtleftR Iright^+leftX_C I-X_L Iright^I sqrtR^+leftX_C-X_Lright^ With the expressions for the reactances X_C and X_L we find Zomega sqrtR^+leftfracomega C-omega Lright^ The phase shift is given by see figure tanDeltaphi fracV_L-V_CV_R fracX_L I-X_C IR I fracX_L-X_CR fracomega L-fracomega CR Alternatively the expressions can be derived using the complex reactances tildeX_L and tildeX_C. For a series circuit the total complex impedance corresponds to the of the partial values: tildeZ R+tildeX_L+tildeX_C R+jomega L+fracjomega C R+jomega L-fracjomega C R+jleftomega L-fracomega Cright The real impedance is given by the modulus of the complex impedance: Z sqrtRetildeZ^+ImtildeZ^ sqrtR^+leftomega L-fracomega Cright^ The phase shift corresponds to the argument: tanDeltaphi fracImtildeZRetildeZ fracomega L-fracomega CR