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Why is the impedance in the parallel resonant circuit maximum at the resonant frequency?

For example if we have two resistors in parallel the equivalent resistance is less than the values of the resistors in the circuit. Isn't this similar to the LC parallel circuit?

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    \$\begingroup\$ In a parallel resonant circuit, charge swings from inductor to capacitor and back. In an ideal circuit no charge leaks away (current). It is important to realize that this only happens when the circuit is at its resonance frequency. It acts as a tank, it slowly fills (which does cause current!) up and once the charge in the circuit causes the same voltage swing as the source voltage, the charge just swings up and down and back again. \$\endgroup\$ – jippie Jun 28 '15 at 17:14
  • \$\begingroup\$ But how does that mean the impedance is high ? @jippie \$\endgroup\$ – jht Jun 28 '15 at 17:18
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    \$\begingroup\$ See the formulas here: allaboutcircuits.com/textbook/alternating-current/chpt-6/… \$\endgroup\$ – Cornelius Jun 28 '15 at 17:21
  • \$\begingroup\$ The basic concept that the sum of Impedences goes to zero holds in a Series or Parallel Resonant Circuit. Its just that in parallel resonant circuit , the components are are connected in Parallel, the total impedence goes to infinity. (Suggested by Andy in Answer). \$\endgroup\$ – sundar Sep 16 '17 at 15:43
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All paralleled components produce an impedance that is "product" divided by "sum" and, at resonance, the "sum" part becomes zero because X\$_L\$ = -X\$_C\$. This means infinite impedance because "something" divided by "zero" produces infinity.

The impedances are opposite but equal in magnitude at resonance and, because they share the same voltage, the currents will be equal in magnitude but of opposite sign. The inductive current lags the voltage by 90 degrees and the capacitive current leads the voltage by 90 degrees hence, the net current is zero in the steady-state situation.

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    \$\begingroup\$ Your answer would considerably improve if you added the formula you refer to. It took me couple moments to figure out what you are writing about. \$\endgroup\$ – jippie Jun 28 '15 at 18:01
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Mathematically- do the math.

Intuitively- current circulates between the reactive parts (capacitor and inductor) rather than going in or out of the terminals of the circuit. It takes a while to build up if the Q of the circuit is high. The energy sloshes back and forth inside the circuit- kind of a 'tank'.

schematic

simulate this circuit – Schematic created using CircuitLab

Voltage across L-C:

enter image description here

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Other answers focused on the concept, I'll focus on the math.

You ask if it is similar to the parallel of two resistors, the answer is yes, but the formula you use when dealing with impedances needs complex numbers and that is the catch.

If two two-terminal elements of impedances \$Z_1\$ and \$Z_2\$ are in parallel they are equivalent to a single impedance whose value is calculated using the same formula you use for paralleled resistors:

\[ Z_{eq} = \dfrac{1}{\dfrac 1 Z_1 + \dfrac 1 Z_2} = \dfrac {Z_1 \cdot Z_2}{Z_1 + Z_2} \]

the problem is that if the two impedances are complex numbers.

Let's consider the case you are interested in: a parallel resonant circuit, that is a capacitor in parallel with an inductor, where I'll represent the resistive components as resistor in parallel to those two elements (to simplify math, otherwise you could represent the series resistance of the inductor and the ESR of the capacitor with two resistors in series to the reactive elements, if you want). In this case you get:

\[ Z_1 = \dfrac{1}{j 2 \pi C f} \qquad Z_2 = j 2 \pi L f \qquad Z_3 = R \]

Their equivalent impedance is:

\[ Z_{eq} = \dfrac{1}{\dfrac 1 Z_1 + \dfrac 1 Z_2 + \dfrac 1 Z_2 } = \dfrac{1}{j 2 \pi C f + \dfrac{1}{j 2 \pi L f} + \dfrac 1 R } \]

To simplify the math involved I'll use the equivalent admittance \$Y_{eq} = \dfrac 1 Z_{eq}\$ and show you that at resonance its magnitude is minimum, which means the magnitude of the impedance is maximum.

Hence:

\[ Y_{eq} = j 2 \pi C f + \dfrac{1}{j 2 \pi L f} + \dfrac 1 R = \dfrac{(j 2 \pi C f)(j 2 \pi R L f) + R + j 2 \pi L f}{j 2 \pi R L f} = \dfrac {R - 2 \pi R L C f^2 + j 2 \pi L f} {j 2 \pi R L f} \]

Note that minimizing \$ \left| Y_{eq} \right|^2 \$ is equivalent to minimizing \$ \left| Y_{eq} \right| \$ (a positive quantity is minimum iff its square is minimum). Therefore, knowing that the squared magnitude of a ratio is the ratio of the squared magnitudes you get:

\[ \left| Y_{eq} \right|^2 = \dfrac {(R - 2 \pi R L C f^2)^2 + (2 \pi L f)^2} {(2 \pi R L f)^2} \]

Without messing with derivatives, it is clear that that fraction is minimum when the first squared term at numerator drops to zero and this happens when:

\[ R - 2 \pi R L C f^2 = 0 \quad \Leftrightarrow \quad f = \dfrac 1 {\sqrt{ 2 \pi L C } } = f_r \]

where \$f_r\$ is the resonant frequency of the circuit.

Substituting \$ f = f_r\$ into the expression of \$\left|Y_{eq}\right|\$ and doing a bit of simple math you get:

\[ \left| Z_{eq(resonance)} \right| = \dfrac 1 {\left| Y_{eq(resonance)} \right|} = R \]

To summarize, for a parallel resonant circuit the magnitude of the impedance is maximum at the resonant frequency \$f_r\$ and its value is equal to the resistive part \$R\$ of the impedance.

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