A calculator to calculate the equivalent impedance of a resistor, a capacitor and and inductor in parallel. The calculator gives the impedance as a complex numbers in standard form , its modulus and argument which may be used to write the impedance in exponential and polar forms.
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We first give the formulas used in the parallel RLC calculator and the proof of these formulas is presented in the bottom part of the page.
Let
\( Z_R = R \) , \( Z_C = \dfrac{1}{j \omega C} \) , \( Z_L = j \omega L\)
Apply the rule of impedances of a parallel circuits to find the equivalent impedance \( Z \) as follows
\( \dfrac{1}{Z} = \dfrac{1}{Z_R} + \dfrac{1}{Z_C} + \dfrac{1}{Z_L} \)
\( = \dfrac{1}{R} + \dfrac{1}{\dfrac{1}{j \omega C}} + \dfrac{1}{j \omega L} \)
Let
\( X_L = \omega L \) and \( X_C = \dfrac{1}{\omega C} \)
and rewrite the above as
\( \dfrac{1}{Z} = \dfrac{1}{R} + \dfrac{1}{\dfrac{X_C}{j}} + \dfrac{1}{j X_L} \)
\( \dfrac{1}{Z} = \dfrac{1}{R} + \dfrac{j}{{X_C}} - j \dfrac{1}{ X_L} \)
\( \dfrac{1}{Z} = \dfrac{1}{R} + j (\dfrac{1}{{X_C}} - \dfrac{1}{ X_L} ) \)
The Modulus \( \rho \) of the above complex number is given by
\( \rho = \sqrt { \left(\dfrac{1}{R}\right)^2 + \left(\dfrac{1}{{X_C}} - \dfrac{1}{ X_L} \right)^2} \)
and its argument \( \alpha \) is given by
\( \alpha = \arctan \left(\dfrac{\dfrac{1}{{X_C}} - \dfrac{1}{ X_L}}{\dfrac{1}{R}} \right) \)
rearrange
\( \alpha = \arctan \left(\dfrac{R}{X_C}-\dfrac{R}{X_L} \right) \)
We now use the exponential form of complex number to write
\( \dfrac{1}{Z} = \rho e^{j\alpha} \)
We now write the equivalent impedance \( Z \) as a complex number in exponential form by taking the reciprocal of the above
\( Z = \dfrac{1}{\rho} e^{-j \alpha} \)
Writing \( Z \) as \( Z = r e^{j\theta} \), we have
the modulus of \( Z \) as
\( r = 1/\rho = \dfrac{1}{\sqrt { \left(\dfrac{1}{R}\right)^2 + \left(\dfrac{1}{{X_C}} - \dfrac{1}{ X_L} \right)^2}} \)
and the argument of \( Z \) as
\( \theta = \arctan \left(\dfrac{R}{X_L}-\dfrac{R}{X_C} \right) \)
\( f = 1.5 \; kHz \) , \( C = 15 \; \mu F \) , \( L = 20 \; mH \) and \( R = 50 \; \Omega \)
\( X_L = \omega L = 2 \pi f L = 2 \pi 1.5 \times 10^3 \times 20 10^{-3 } = 188.50 \)
\( X_C = \dfrac{1}{\omega C} = \dfrac{1}{ 2\pi f C} = \dfrac{1}{ 2\pi 1.5 \times 10^3 \times 15 10^{-6}} = 7.07\)
Modulus: \( \dfrac{1}{\sqrt { \left(\dfrac{1}{R}\right)^2 + \left(\dfrac{1}{{X_C}} - \dfrac{1}{ X_L} \right)^2}} \)
\( = \dfrac{1}{\sqrt { \left(\dfrac{1}{50}\right)^2 + \left(\dfrac{1}{{7.07}} - \dfrac{1}{ 188.50} \right)^2}} \)
\( = 7.27 \)
Argument: \( \arctan \left(\dfrac{R}{X_L}-\dfrac{R}{X_C} \right) \)
\( = \arctan \left(\dfrac{50}{188.50}-\dfrac{50}{7.07} \right) \)
\( = - 81.64^{\circ} \)
You may input the given values in the calculator and check the results.