USING IMPEDANCE AND ADMITTANCE

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PASSIVE COMPONENTS IN AC CIRCUITS

VOLTAGE AND CURRENT DIVISION

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.

Example 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 e^{– 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 Label command to add the values and the Line 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_eq, can be considered as a complex resultant vector derived using the parallelogram rule from the complex impedances Z_R and Z_{L .}

Example 2

Find the equivalent impedance and admittance of this parallel circuit:

The admittance:

The impedance using the Z_tot= Z₁ Z₂ / (Z₁ + Z₂ ) formula for parallel impedances:

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

Example 3

Find the equivalent impedance of this parallel circuit. It uses the same elements as in Example 1:
R = 12 ohm and L = 10 mH, at f = 159 Hz frequency.

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

Y_eq = Y_R + Y_L = 1/R + 1/ (j*2*p*f*L) = 1/12 – j /10 = 0.0833 – j 0.1 = 0.13 e^{–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:

Example 4

Find the impedance of a series circuit with R = 10 ohm, C = 4 mF, and L = 0.3 mH, at an angular frequency w = 50 krad/s (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

The phasor diagram as generated by TINA

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 to the tip of Z_L. Then we move the tail of Z_C to the tip of Z_R. Now the resultant Z_eq will exactly close the polygon starting from the tail of the first Z_R phasor and ending at the tip of 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 inductive circuit–at least at this frequency!

Example 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 inductive, 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 mH

Example 6

Find the impedance of three components connected in parallel: R = 4 ohm, C = 4 mF, and L = 0.3 mH, at an angular frequency w = 50 krad/s (f = 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 e^{–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 capacitive circuit.

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.

Example 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

hence

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

Example 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. Then, seeing that Z_RC is in parallel with the series-connected L3 and R3, we’ll compute the equivalent impedance of their parallel connection, Z₂. Finally, we calculate Z_eq as the sum of Z₁ and 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 e^–j31.8° ohm

according to TINA’s result.

PASSIVE COMPONENTS IN AC CIRCUITS

VOLTAGE AND CURRENT DIVISION