How to calculate impedance

In this short podcast episode, Bryan explains the science behind adiabatic cooling. Adiabatic cooling occurs in specific HVAC/R applications and in our environment as air temperatures and pressures change.

When we think of cooling, we refer to the loss of heat; we are either referring to the change in the total BTU content of the air mass or the temperature change. Adiabatic cooling takes sensible heat and transforms it into latent heat.

The most simple forms of adiabatic cooling can be seen in cooling towers and swamp coolers. In evaporative or swamp coolers, you have a pad saturated with water, and air moves over it. When air moves over the media, some of the energy helps evaporate the moisture on the pads, so the air loses sensible heat and becomes cooler. The thermal enthalpy (total heat content) stays the same, but some of the sensible heat has transferred to latent heat.

Air that goes through a swamp cooler goes in with higher temperature and lower humidity, and it leaves with a lower temperature and higher humidity. The BTU content stays the same; the energy merely transforms. As a result, we usually only use swamp coolers in arid environments where higher humidity is desirable. You also can’t compare these to compression-refrigeration systems because compression refrigeration aims to change the BTU content and is NOT adiabatic.

When we run air over an evaporator coil, some of the water vapor in the air condenses to liquid water in the drain pan. Some of the energy in the refrigerant changes the state of the water vapor to liquid water instead of changing the temperature. You’ll see a lower delta T when your return relative humidity (RH) is higher.

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Examples on how to use the rules of impedances connected in series and parallel to calculate equivalent impedances in various AC circuits and present them as complex numbers in standard, complex and polar forms. Detailed solutions to the examples are also presented.

Examples with Solutions

Example 1
Find the equivalent impedance between points A and B in the circuit given below and write it in exponential and polar form. .

Solution to Example 1
Let $$Z_1$$ be the impedance of resistor R and hence $$Z_1 = R$$
Let $$Z_2$$ be the impedance of the capacitor $$C$$ and the inductor $$L$$ that are in parallel.
$$Z_1$$ and $$Z_2$$ are in series and the equivalent impedance $$Z_$$ is given by the rule of series impedances by
$$Z_ = Z_1 + Z_2$$

The impedance of a capacitor with capacitance $$C$$ in complex form is equal to $$\dfrac<1> < j \omega C>$$
The impedance of an inductor with inductance $$L$$ in complex form is equal to $$j \omega L$$
We now use the rule of parallel impedances to calculate $$Z_2$$ as
$$\dfrac<1> = \dfrac<1> + \dfrac<1><\dfrac<1>< j \omega C>>$$
which may be written as
$$\dfrac<1> = \dfrac<1> + j \omega C$$
Write the right hand side with a common denominator
$$\dfrac<1> = \dfrac<1-\omega^2 C L>$$
Solve for $$Z_2$$
$$Z_2 = \dfrac < 1-\omega^2 C L>$$
Substitute $$Z_1$$ and $$Z_2$$ by their expressions to obtain $$Z_$$
$$Z_ = R + \dfrac < 1-\omega^2 C L>$$
Find the modulus $$|Z_|$$ and argument $$\theta$$ of $$Z_$$
$$|Z_| = \sqrt < 1-\omega^2 C L )>\right)^2 >$$
$$\theta = \arctan \dfrac<\omega L>$$
In exponential form, the equivalent impedance is given by
$$Z_ = \sqrt < 1-\omega^2 C L )>\right)^2 > e^<\arctan \dfrac<\omega L>>$$
In polar form it is written as
$$Z_ = \sqrt < 1-\omega^2 C L )>\right)^2 > \; \angle \; <\arctan \dfrac<\omega L>>$$

Example 2
Find the equivalent impedance between points A and B in the circuit given below and write it in exponential and polar forms given:
$$L_1 = 20 \; mH$$ , $$C_1 = 10 \; \mu F$$ , $$L_2 = 40 \; mH$$ , $$C_2 = 30 \; \mu F$$ the frequency of the signal $$f = 1.5 \; kHz$$

Solution to Example 2
Let $$Z_1$$ be the impedance of the capacitor $$C_1$$ and the inductor $$L_1$$ that are in parallel.
Let $$Z_2$$ be the impedance of the capacitor $$C_2$$ and the inductor $$L_2$$ that are in parallel.
$$Z_1$$ and $$Z_2$$ are in series as shown below hence using We now use the rule of series impedances to calculate $$Z_$$ as follows
$$Z_ = Z_1 + Z_2$$

We now use the rule of parallel impedances to calculate $$Z_1$$ and $$Z_2$$ as follows
$$\dfrac<1> = \dfrac<1> + \dfrac<1><\dfrac<1>< j \omega C_1>>$$
rewrite the above as
$$\dfrac<1> = \dfrac<1> + j \omega C_1$$
write the right side with common denominator
$$\dfrac<1> = \dfrac<1 - \omega^2 L_1 C_1>$$
Solve for $$Z_1$$
$$Z_1 = \dfrac <1 - \omega^2 L_1 C_1>$$
$$Z_2$$ may be calculated in a similar way as $$Z_1$$ and is given by
$$Z_2 = \dfrac <1 - \omega^2 L_2 C_2>$$
We now substitute $$Z_1$$ and $$Z_2$$ by their expression in $$Z_$$ to obtain
$$Z_ = \dfrac <1 - \omega^2 L_1 C_1>+ \dfrac <1 - \omega^2 L_2 C_2>$$
Fcator $$j \omega$$ out and rewrite $$Z_$$ as
$$Z_ = j \omega \left (\dfrac< L_1> <1 - \omega^2 L_1 C_1>+ \dfrac< L_2> <1 - \omega^2 L_2 C_2>\right)$$
Substitute by the numerical values of $$L_1 , C_1 , L_2 , C_2$$ and $$\omega = 2 \pi f$$ into $$Z_$$
$$Z_ \approx – j 14.81$$
Note that $$Z_$$ is purte imaginary and therefore the modulus $$|Z_|$$ and argument $$\theta$$ of $$Z_$$ are given by
$$|Z_| \approx 14.81$$
$$\theta = – \pi / 2$$
Exponential form
$$Z \approx 14.81 \; e^ <-j \pi/2>$$
Polar form
$$Z \approx 14.81 \angle – \pi/2$$

Example 3
Find the equivalent impedance between points A and B in the circuit given below and write it in exponential and polar forms given that
$$R_1 = 20 \; \Omega$$ , $$C_1 = 50 \; \mu F$$ , $$C_2 = 40 \; \mu F$$ , $$R_2 = 80 \; \Omega$$ the frequency of the signal $$f = 0.5 \; kHz$$

Solution to Example 3
$$Z_1 = R_1$$
$$Z_2 = \dfrac<1>$$
$$Z_3 = R_2 + \dfrac<1>$$
$$Z_2$$ and $$Z_3$$ are in pararllel and their equivalent impedance $$Z_ <2,3>$$ using the rule of parallel impedances is given by
$$\dfrac<1>> = \dfrac<1> + \dfrac<1>$$
$$Z_ <2,3>= \dfrac$$

$$Z_1$$ and $$Z_ <2,3>$$ are in series, hence
$$Z_ = Z_1 + Z_ <2,3>= Z_1 + \dfrac$$
Substitute
$$Z_ = R_1 + \dfrac<\dfrac<1> \cdot (R_2 + \dfrac<1>)><\dfrac<1> + R_2 + \dfrac<1>>$$
Substitute by the numerical values of $$R_1 = 20 \; \Omega$$ , $$C_1 = 50 \; \mu F$$ , $$C_2 = 40 \; \mu F$$ , $$R_2 = 80 \; \Omega$$ the frequency of the signal $$f = 0.5 \; kHz$$ to obtain
$$Z_ \approx 20.49 -6.29 j$$
Modulus of $$Z_$$
$$| Z_ | \approx \sqrt <20.49^2 + (-6.29)^2 >= 21.43$$
Argument of $$Z_$$
$$\theta \approx \arctan (\dfrac<-6.29><20.49>) = -0.20 rad$$ or $$\theta = -17.07^ <\circ>$$
Hence in exponential form
$$Z_ \approx 21.43 e^ < -0.20 j>$$
and in polar form
$$Z_ \approx 21.43 \angle -17.07^ <\circ>$$

How do you calculate the impedance of a transformer?

Effective Percent Impedance

1. Transformer reactance Xt = (kV2/MVA) x %Z/100 = (0.482 / 0.5) x 0.06 = 0.027648 ohms.
2. Rated secondary current = 500,000 / (480 x 1.732) = 601.4 amps.
3. Actual Load current = 300 amps.
4. Voltage drop at actual load = 300 x 1.732 x 0.027648 = 14.36 volts (14.36 / 480 = 0.0299, or 3% of 480 volts)

What is the purpose of open circuit test?

The open-circuit test, or no-load test, is one of the methods used in electrical engineering to determine the no-load impedance in the excitation branch of a transformer. The no load is represented by the open circuit, which is represented on the right side of the figure as the “hole” or incomplete part of the circuit.

What is open and short circuit test?

The open circuit and short circuit test are performed for determining the parameter of the transformer like their efficiency, voltage regulation, circuit constant etc. These tests are performed without the actual loading and because of this reason the very less power is required for the test.

Why no load and impedance test is done in transformer?

Open circuit test or no load test on a transformer is performed to determine ‘no load loss (core loss)’ and ‘no load current I0’. The circuit diagram for open circuit test is shown in the figure below. Usually high voltage (HV) winding is kept open and the low voltage (LV) winding is connected to its normal supply.

What is the impedance of a transformer?

The per-unit impedance describes that percentage of the rated voltage required to produce full load current while the transformer output is shorted. The lower the impedance, the lower the voltage required to produce full load current.

What is per unit impedance of transformer?

The per unit system of calculation is a method whereby system impedances and quantities are normalized across different voltage levels to a common base. Any per unit impedance will have the same value on both the primary and secondary of a transformer and is independent of voltage level.

Why do we use short-circuit test in transformer?

Hence the short-circuit test of a transformer is used to determine copper losses in the transformer at full load. It is also used to obtain the parameters to approximate the equivalent circuit of a transformer.

What is difference between open circuit and short circuit?

The open circuit voltage is the voltage difference measured between two terminals when no current is drawn or supplied. The short circuit current is the current that flows when the terminals are forced to have zero voltage difference.

Why is a transformer rated in kVA?

Transformers are rated in kVA because the losses occurring in the transformers are independent of power factor. KVA is the unit of apparent power. It is a combination of real power and reactive power. Transformers are manufactured without considering the load being connected.

Can a transformer work on direct current?

Transformers do not pass direct current (DC), and can be used to take the DC voltage (the constant voltage) out of a signal while keeping the part that changes (the AC voltage). In the electrical grid transformers are key to changing the voltages to reduce how much energy is lost in electrical transmission.

What is impedance and why is it important?

Whether you are working with digital or analog signals, you’ll most likely need to match impedances between a source, transmission line, and load. The reason impedance matching is important in a transmission line is to ensure that a 5 V signal sent down the line is seen as a 5 V signal at the receiver.

How transformer is used in impedance matching?

The simplest way to match load impedance in AC circuits is to use a transformer — a highly efficient device that transfers electrical energy from one circuit to another through electromagnetic induction: Two coils of wire (windings) are wrapped around an iron or ferrite core, and when the primary winding experiences …

What is per unit system and its advantages?

per unit (pu) system advantages: Circuits are simplified. Voltages have same range in per unit in all parts of the system from EHV system to distribution and utilization. By choosing suitable base kV for the circuits the per unit reactance remains the same, referred to either side of the transformer.

What is the other name of short-circuit test?

In Short Circuit (SC) test, the primary or HV winding is connected to the AC supply source through voltmeter, ammeter, wattmeter and a variac as shown in figure. This test is also called as Reduced Voltage Test or Low Voltage Test.

Enter the impedance of each individual speaker and select whether they are connected in series or parallel. The speaker impedance calculator will display the equivalent total impedance of the setup.

Speaker Impedance Formula

The following is the speaker impedance formula:

Parallel: I = 1 / [ (1/X1) + (1/X2) + ( 1/X3) … ]

Series: I = X1 + X2 + X3 ….

• Where I is the equivalent impedance of the speaker setup (ohms)
• X1,X2, etc are the individual impedance of each speaker (ohms)

Speaker Impedance Definition

A speaker impedance is defined as the total impedance of the electrical circuit connecting 1 or more speakers. Impedance is a measure of the resistance to an alternating current. Impedance is a combination of typical ohmic resistance and reactance.

Speaker Impedance Importance

Why is speaker impedance important? A speaker impedance is a key factor in the power required to achieve a certain signal level. This can be seen if you look at the formula for power, P = V^2 / Z, where P is the power, V is the voltage, and Z is the impedance. When an impedance is lower, the required power will be higher, and when the impedance is higher the required power will be low.

Therefore, a high impedance system will be less taxing on an amplifier when trying to achieve the same voltage.

How to calculate speaker impedance?

Example Problem:

First, determine whether the speaker is in parallel or series. For this example, the speaker is in parallel.

Next, determine the impedance of each individual component. For this problem, these are 4, 12, and 14 respectively.

Finally, calculate the speaker impedance using the formula above:

Series: I = X1 + X2 + X3 ….

Series: I = 4 + 12 + 14

Series: I = 30 ohms

A speaker impedance typically only affects the total voltage that is required to meet a certain power level. So, if there is a proper amplifier the impedance should not affect the volume.

Yes, you can have more than 1 speaker with varying impedances, but it should be noted that you should have an amplifier capable of handling the equivalent impedance.

I want to find the impedance in this circuit.
Is it possible to calculate the capacitor as 1 / jwc after removing the voltage from this circuit?
I am not sure because it is confusing.
Thank you for letting us know how to do the calculations with and without the dashed part of the circuit.

• Apr 14, 2020
• #2

KlausST

Super Moderator

What do you mean by "removing the voltage"?

Do you have to calculate it in general, or for a (not) given frequency and waveform?

(A forum is not meant that others do your job/homework, but we will help to rectify your mistakes. Thus
Show your ideas, show what you have done so fare. Show your calculations.

• Apr 14, 2020
• #3

kareno

Newbie level 4

It was my first time, so I was not good at posting. Sorry.

Eliminating the voltage meant that I could ignore the voltage when calculating the impedance.

When I calculated it my way, it came out like a picture.
If I use w instead of frequency, I am wondering this is correct.

• Apr 14, 2020
• #4

KlausST

Super Moderator

"w" still includes frequency, since w = 2 x Pi x f

I still don’t understand. How is voltage involved in impedance calculation?

• Apr 14, 2020
• #5

kareno

Newbie level 4

I am wondering how to handle the voltage.
The equation posted in my comment above is to find the impedance using only resistors and capacitors. is this right?
If not, please let me know how to solve it.

• Apr 14, 2020
• #6

• Apr 14, 2020
• #7

kareno

Newbie level 4

oh then can i consider voltage as short circuit? right?
then is my calculation right?

• Apr 14, 2020
• #8

c_mitra

We understand. But once you have made a schematic, you have progressed a lot.

Now consider the impedance just like an AC resistance.

You apply a voltage (AC at some given frequency) and you measure the corresponding current (you need to measure again the AC current).

Yes, you need to define two points in the above circuit where you will apply the AC voltage. So you need to measure the AC impedance between two given points in the diagram.

Just like the DC resistance of a circuit depends on the points you select to put the test probes (the two pointed things in your multimeter) on.

Fortunately your circuit has only resistors and capacitors. Voltage sources contribute zero impedance to the overall circuit.

Yes, the impedance of the capacitor is 1/(jwc) and the rest is simply series and parallel combinations.

Resistance and impedance both represent opposition to electric current. However, resistance opposes both direct and alternating current, while the reactance component of impedance opposes only changing current. Calculations for DC circuits can be done with scalar quantities and ordinary algebra. But impedance is a phasor quantity in AC circuits, and so calculations for impedance networks are based on phasor algebra.

With phasor algebra, all the relationships for resistance networks also apply to impedance networks.

Impedances in Series

At any particular instant, the circuit relationships in Figure 1(a) are exactly the same as in a DC circuit that has the voltages and currents that prevail at that instant. Consequently, Kirchhoff’s voltage law gives

To extend Kirchhoff’s voltage law to the much more useful RMS values for AC circuit parameters, we must use phasor quantities. Then, the Kirchhoff’s voltage-law relationship for Figure 1(a) can be written as

Each term in this equation is a phasor, and the addition must be done with phasor algebra.

Figure 1 Kirchhoff’s voltage law applied to an AC circuit

We can apply the phasor form of Kirchhoff’s voltage law to the series circuit of Figure 1(b):

Dividing every term in Equation 1 by the common current in the series circuit gives

By definition, Z = V/I. Therefore,

The total impedance of impedances in series is their phasor sum. This relationship can be extended to any number of impedances in series:

Example 1

Find the total impedance when $<_<1>>=60\angle +<<60>^>\Omega \text< and ><_<2>>=80\angle –<<45>^>\Omega$ are connected in series. Give the answer in polar form.

Solution

Step 1

Draw a circuit diagram and sketch an impedance diagram to indicate the approximate total impedance, as shown in Figure 2 .

Step 2

Convert the impedances to rectangular form, and add the components:

Figure 2 Impedances in series

Step 3

Convert the total impedance into polar form:

Example 1a

Find the total impedance when $<_<1>>=10,000+j15,000\Omega \text< , ><_<2>>=47,000+j0\Omega \text< and ><_<3>>=25,000-j10,000\Omega$ are connected in series.

Solution

It may appear from the simple solution of Example 1a that impedance should always be expressed in rectangular coordinates. However, the rectangular coordinates of an impedance represent resistance and a reactance in series. Therefore, the addition process of Example 1a applies only to series impedances.

We can now extend the voltage-divider principle to series impedances. With phasor quantities, we can write the following expression:

Example 2

Find the voltage drop across Z1 in the circuit of Figure 2(a) when the applied voltage is $120\angle <<0>^>$.

Solution

Step 1

Use Ohm’s law with ZT as determined in Example 1:

Impedances in Parallel

As with parallel DC circuits, we can analyze parallel AC circuits by considering the circuit’s ability to conduct current. Two sets of parameters are proportional to a circuit’s ability to pass current:

• The total of all branch currents

• The total of all branch admittances

For parallel branches in an AC circuit, Kirchhoff’s current law gives

Dividing every term in Equation 4 by the common voltage across parallel branches gives

Since admittance is defined as Y = 1/Z,

The total admittance of parallel branches is the phasor sum of the branch admittances.

The equivalent impedance of a parallel AC circuit can be determined in two ways: by the total-current method (from Equation 4),

And by the total-admittance method (from Equation 5)

For the special case of two branches in parallel,

Inverting this equation gives

Example 3

Find the equivalent impedance for $<_<1>>=60\angle +<<60>^>\Omega \text< and ><_<2>>=80\angle –<<45>^>\Omega$ connected in series.

Solution

Step 1

Draw a schematic diagram and a current phasor diagram, as in Figure 3 .

Figure 3 Total-current method for Example 3

Step 2

Assume a convenient value of applied voltage, such as E = 240 ∠ 0° V. Then determine the branch currents:

Step 3

Convert the branch currents to rectangular form and add the components:

Step 4

Convert the total current to polar form:

Step 5

Use the total current to calculate the equivalent impedance:

Since the same voltage appears across parallel branches, we can establish the current-divider principle for AC circuits from the relationship:

Therefore, the ratio of the currents in any two parallel branches of an AC circuit is the inverse of the ratio of the impedances of the two branches. However, it is more convenient to express the current-divider principle in terms of branch admittances. Since,

As with DC circuits, the current-divider equation 10 is the dual of the voltage-divider equation 3. When there are only two parallel branches, we can use Equation 8 to substitute for 1/YT in Equation 10:

Example 4

What current does a 10∠-45°kΩ load draw from the Norton-equivalent source shown in Figure 4 ?

While Ohm’s Law applies directly to resistors in DC or in AC circuits, the form of the current-voltage relationship in AC circuits in general is modified to the form:

where I and V are the rms or “effective” values. The quantity Z is called impedance. For a pure resistor, Z = R. Because the phase affects the impedance and because the contributions of capacitors and inductors differ in phase from resistive components by 90 degrees, a process like vector addition (phasors) is used to develop expressions for impedance. More general is the complex impedance method.

Combining impedances has similarities to the combining of resistors, but the phase relationships make it practically necessary to use the complex impedance method for carrying out the operations. Combining series impedances is straightforward:

Combining parallel impedances is more difficult and shows the power of the complex impedance approach. The expressions must be rationalized and are lengthy algebraic forms.

The complex impedance of the parallel circuit takes the form

when rationalized, and the components have the form

Impedances may be combined using the complex impedance method.

The units for all quantities are ohms. A negative phase angle implies that the impedance is capacitive, and a positive phase angle implies net inductive behavior.

Calculating total impedance Z in electrical AC circuits. In collection of examples you can find derivation of formulas for capacitive reactance XC and inductive reactance XL. Impedance is wider magnitude than resistance. Impedance has three components which are: resistance, capacitive reactance and inductive reactance. Impedance in general case is a vector. Components of impedance are geometric sum, because they also are vectors. Complex numbers are applied to calculate impedance and its components.

Components of impedance

Impedance has three components. In general impedance is a vector which is built of complex number.
Z=R+j·XL-j·XC
where
• R=ρ·l/S – resistance [Ω]
• XL=ω·L – inductive reactance [Ω]
• XC=1/(ω·C) – capacitive reactance [Ω]

Total impedance determination – example 1

Designation of total impedance and its components for AC circuit. In example circuit impedance Z which is seen from power supply terminals will be calculated. Circuit is built from two resistors R, capacitor C and inductivity L. We will plot also vector diagram of currents and voltages in circuit.

Total impedance determination – example 2

Designation of total impedance Z for electrical AC circuit. Considered electrical circuit built with three resistors R, capacitor C and inductivity L. All parts in the circuit have the same value of impedance which is 1[Ω]. Parts in circuit are connected in mixed ways because there are serial connections and parallel connections. Total impedance is designated as impedance which is seen from power terminals. In example relation between impedance Z and admittance Y will be applied → Y=1/Z.
Z=R+j·X → Z=R+j·(XL-XC)
Y=G+j·B

Total impedance determination – example 3

Designation of total impedance and its components for AC circuit. In example circuit impedance which is seen from power supply terminals will be calculated. Circuit is built from two resistors, two coils and one capacitor. Formulas for Ohm law in AC circuit will be also written.

Total impedance determination – example 4

Designation of total impedance and its components for AC circuit. In example circuit impedance which is seen from power supply terminals will be calculated. Circuit is built from three coils, three capacitors and one resistor. Ohm law in AC circuit will be also written.

Total impedance determination – example 5

Designation of total impedance Z and its components for AC circuit. In example circuit impedance which is seen from power supply terminals will be calculated. Circuit is built with two capacitors, inductivity and resistor. Impedances of every part in circuit is known. Circuit has mixed connections, parallel and serial . Relation between impedance Z and admittance Y will be applied → Y=1/Z.

Total impedance determination – example 6

Designation of total impedance Z and its components for AC circuit. In example circuit impedance which is seen from power supply terminals will be calculated. Circuit is built with two inductivities, three resistors and capacitor. Impedances of every part in circuit is known. Circuit has mixed connections, parallel and serial . Relation between impedance Z and admittance Y will be applied → Y=1/Z. Admittance is usually used to calculate impedance of parallel connection between parts.

Total impedance determination – example 7

Electrical circuit is built with three passive parts, resistor R1=2[Ω], capacitor C1=2[μF] and inductivity L1=100[mH]. All parts in circuit are connected in parallel. Circuit is supplied with voltage U. Voltage’s rms value is 230[VAC]. Voltage has frequency f=50[Hz]. Subject of example is calculation of current I, which flows through circuit. In example total impedance Z will be calculated. Ohm’s law for AC circuits will be applied to calculate current II=U/Z. All calculations in example are made with application of complex numbers.