Импеданстар ЖАНА кабыл ПАЙДАЛАНУУ

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As we saw in the previous chapter, impedance and admittance can be manipulated using the same rules as are used for DC circuits. In this chapter we will demonstrate these rules by calculating total or equivalent impedance for series, parallel and series-parallel AC circuits.

мисал 1

Find the equivalent impedance of the following circuit:

R = 12 ohm, L = 10 mH, f = 159 Hz

The elements are in series, so we realise that their complex impedances should be added:

Z_eq = Z_R + Z_L = R + j w L = 12 + j* 2 *p* * 159 0.01 = (12 + j 9.99) ohm = 15.6 e^j39.8° Ohm.

Y_eq = 1 /Z_eq = 0.064 д^{- j 39.8°} S = 0.0492 - j 0.0409 S

We can illustrate this result using impedance meters and the Phasor Diagram in
TINA v6. Since TINA’s impedance meter is an active device and we are going to use two of them, we must arrange the circuit so that the meters don’t influence each other.
We have created another circuit just for the measurement of the part impedances. In this circuit, the two meters do not “see” each other’s impedance.

The Analysis/AC Analysis/Phasor diagram command will draw the three phasors on one diagram. We used the Auto этикеткасы command to add the values and the сызык command of the Diagram Editor to add the dashed auxiliary lines for the parallelogram rule.

The circuit for measuring the impedances of the parts

As the diagram shows, the total impedance, Z_.ж, татаал пайда багыты аркылуу алынган катары кароого болот parallelogram rule татаал күчөткүчтөр чейин Z_R жана Z_{L .}

мисал 2

Find the equivalent impedance and admittance of this parallel circuit:

The admittance:

колдонуу импеданстар Z_{Чонбури}= Z₁ Z₂ / (Z₁ + Z₂ ) параллелдүү күчөткүчтөр үчүн формула:

ТИНА бул маселени чече алабы дагы бир жолу анын котормочусу менен:

мисал 3

Find the equivalent impedance of this parallel circuit. It uses the same elements as in Example 1:
R = 12 Ohm жана L = 10 М.Х., п = 159 Hz жыштыгы.

For parallel circuits, it’s often easier to calculate the admittance first:

Y_eq = Y_R + Y_L = 1 / R + 1 / (j*2*p*е * L) = 1 / 12 - j /10 = 0.0833 – j 0.1 = 0.13 д^{-j 50°} S

Z_eq = 1 / Y_eq = 7.68 e ^{j 50°} Ohm.

Another way TINA can solve this problem is with its Interpreter:

мисал 4

Find the impedance of a series circuit with R = 10 ohm, C = 4 mF, жана L = 0.3 М.Х., бурчтук жыштыгы w = 50 krad / с (f = w / 2p = 7.957 KHz).

Z = R + j w L - j / wC = 10 + j 5*10⁴ * * 3 10^-4 - j / (5 * 10⁴ * * 4 10^-6 ) = 10 + j 15 - j 5

The circuit for measuring the impedances of the parts

Тина тарабынан түзүлгөн катары phasor диаграмма

Starting with the phasor diagram above, let’s use the triangle or geometric construction rule to find the equivalent impedance. We start by moving the tail of Z_R учунан Z_L. Ошондо биз, куйрук түрткү Z_C учунан Z_R. Now the resultant Z_eq will exactly close the polygon starting from the tail of the first Z_R phasor жана учундагы бүтөт Z_C.

The phasor diagram showing the geometric construction of Z_eq

Check your calculations using TINA’s Analysis menu Calculate nodal voltages. When you click on the Impedance meter, TINA presents both the impedance and admittance, and gives the results in algebraic and exponential forms.

Since the circuit’s impedance has a positive phase like an inductor, we can call it an тыянак райондук–at least at this frequency!

мисал 5

Find a simpler series network that could replace the series circuit of example 4 (at the given frequency).

We noted in Example 4 that the network is тыянак, so we can replace it by a 4 ohm resistor and a 10 ohm inductive reactance in series:

X_L = 10 = w* L = 50 * 10³ L

® L = 0.2 М.Х.

мисал 6

Find the impedance of three components connected in parallel: R = 4 ohm, C = 4 mF, жана L = 0.3 mH, at an angular frequency w = 50 krad / с (= е w / 2p = 7.947 KHz).

1/Z = 1 / R + 1 / j w L + jwC = 0.25 - j/15 +j0.2 = 0.25 +j 0.1333

Z = 1 / (0.25 + j 0.133) = (0.25 - j 0.133)/0.0802 = 3.11 – j 1.65 = 3.5238 д^{-j 28.1°} OHMS.

The Interpreter calculates phase in radians. If you want phase in degrees, you can convert from radians to degrees by multiplying by 180 and dividing by p. In this last example, you see a simpler way—use the Interpreter’s built in function, radtodeg. There is an inverse function as well, degtorad. Note that this network’s impedance has a negative phase like a capacitor, so we say that—at this frequency—it is a емкостный райондук.

In Example 4 we placed three passive components in series, while in this example we placed the same three elements in parallel. Comparing the equivalent impedances calculated at the same frequency, reveals that they are totally different, even their inductive or capacitive character.

мисал 7

Find a simple series network that could replace the parallel circuit of example 6 (at the given frequency).

This network is capacitive because of the negative phase, so we try to replace it with a series connection of a resistor and a capacitor:

Z_eq = (3.11 - j 1.66) Ohm = R_e -j / wC_e

R_e = 3.11 Ohm w* C = 1 / 1.66 = 0.6024

демек,

R_e = 3.11 Ohm
C = 12.048 mF

You could, of course, replace the parallel circuit with a simpler parallel circuit in both examples

мисал 8

Find the equivalent impedance of the following more complicated circuit at frequency f=50 Hz:

We need a strategy before we begin. First we’ll reduce C and R2 to an equivalent impedance, Z_RC. Андан кийин, бул Корган көрүп_RC is in parallel with the series-connected L3 and R3, we’ll compute the equivalent impedance of their parallel connection, Z₂. Акыр-аягы, биз Корган эсептеп_eq Z суммасы катары₁ жана Z₂.

Here’s the calculation of Z_RC:

Here’s the calculation of Z₂:

And finally:

Z_eq = Z₁ + Z₂ = (55.47 - j 34.45) Ohm = 65.3 д^-j31.8° Ohm

according to TINA’s result.