Equivalent Impedance In A RLC Circuit

**Strings (S _{i}P_{j}A_{jk}) = S_{7}P_{3}A_{31} Base Sequence = 12735 String Sequence = 12735 - 3 - 31 **

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Equivalent Impedance In A RLC Circuit

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Determine the equivalent impedance of the circuit shown in figure 4.36, given that:

v_{s}(t) = 636cos(3,000t + π/12)

R_{1} = 3.3 kΩ; R_{2} = 22 kΩ

L = 1.90 H; C = 6.8 nF.

**The string**:

S_{7}P_{3}A_{31} (Force - Pull).
**The math**:

Pj Problem of interest is of type *force*. The resistance to flow or motion is a *foce problem* (force - pull).

When the signals from the active elements of a circuit are sinusoidal excitations (e.g. AC circuits), the concept of *resistance* is replaced by the concept of *impedance*. In essence, *impedance* may be viewed as a *complex frequency* in the sense that inductors and capacitors are *frequency-resistors* whose resistances (called *reactance*) are functions of the frequency of the sinusoidal excitation. Nonetheless, the fundamental DC theories (Ohm's and Kirchoff's laws) remain applicable.

R_{1} is in series with L:

So, Z_{1} = R_{1} + jωL

Where jωL is the reactance of the inductor.

So, Z_{1} = 10^{3}(3.3 + j5.7)

Phasor form for Z_{1} :

Z_{1} = 10^{3}(6.59)<1.046.

R_{2} is in series with C:

So, Z_{2} = R_{2} + 1/(jωC)

Where 1/(jωC) is the reactance of the capacitor.

So, Z_{2} = 10^{3}(22 - j49)

Phasor form for Z_{2}:

Z_{2} = 10^{3}(53.71)< -1.1495

Z_{1} + Z_{2} = 25.3 - j43.3.

Phasor form for Z_{1} + Z_{2}:

10^{3}(50.15)<-1.052.

Z_{1} is parallel to Z_{2}:

So, Z_{eq} = (Z_{1})(Z_{2})/(Z_{1} + Z_{2})

So, Z_{eq} = 10^{6}(6.59<1.046)(53.71< -1.1495)/10^{3}(50.15<-1.052)

So, Z_{eq} = 7.05<0.94 kΩ

The *point* **.** is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.

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