Introduction to Electrical Engineering - Basic vocational knowledge (Institut für Berufliche Entwicklung, 213 p.)
 7. Alternating Current
 7.1. Importance and Advantages of Alternating Current 7.2. Characteristics of Alternating Current 7.3. Resistances in an Alternating Current Circuit 7.4. Power of Alternating Current

### 7.3. Resistances in an Alternating Current Circuit

Effective resistance R

Loads which completely convert the electrical energy into heat energy are called effective resistances. They include thermal appliances, incandescent lamps and wire and film resistors used for current limitation. The behaviour of effective resistances in alternating current is the same as in direct current. Ohm’s law dealt with in Section 3.2. is applicable to them without any restriction. Its resistance value R is independent of the frequency of the alternating current (Fig. 7.7.). The voltage has the same phase as the current. In Fig. 7.8., the vector diagram and the line diagram for current and voltage with an effective resistance is represented. Ideal effective resistances, also known as active resistances, have no inductance and no capacity.

Fig. 7.7. Effective (or active) resistance as a function of frequency

Fig. 7.8. Current and voltage curves for an effective resistance Reactance X

Coils and capacitors do not convert the electrical energy into heat energy but store it in a magnetic or electrical field. Such components have a reactance. A distinction is made between inductive reactances and capacitive reactances.

· Inductive reactance XL

When an alternating current flows through a coil, a voltage is induced in the latter which offers a resistance to the passage of current. This capability of offering resistance is the greater, the greater the inductance and the rate of current variation (hence, the frequency) are. Consequently, the coil has a resistance which increases with increasing frequency.

XL = wL = 2pfL

(7.7.)

where

 XL inductive reactance w angular frequency f frequency L inductance

[XL] = 1/s · (V·· s)/A = V/A = W

Fig. 7.9. shows the function XL = f (w)

In Section 5.3.3. proof has been given of the fact that a coil imparts sluggishness to a current. When the current passes through its maximum value (for alternating current this is , its rate of variation has the smallest value and the counter-voltage induced in the coil or the voltage drop is equal to zero. Consequently, there is a phase shift produced between current and voltage of , that is to say, the current lags behind the voltage. Fig. 7.10. shows vector and line diagrams to illustrate these correlations. Ideal coils do not have an effective resistance and no capacity.

Fig. 7.9. Inductive reactance as a function of frequency

Fig. 7.10. Current and voltage curves for an inductive reactance

Capacitive reactance XC,

When an alternating voltage is applied to a capacitor, then a continuously varying charging and discharging current is produced which apparently penetrates the capacitor. This current is the greater, the greater the capacity and the rate of voltage variation (i.e. the frequency) are. Consequently, the capacitor has a resistance which becomes smaller with increasing frequency.

XC = 1/(wC) = 1/(2pfC)

(7.8.)

where:

 XC capacitive reactance w angular frequency f frequency C capacity

Fig. 7.11. shows the function XC = f (w)

Fig. 7.11. Capacitive reactance as a function of frequency

In Section 6.2.2. it has been explained that no sudden voltage changes are possible in a capacitor. First a current must flow before a voltage can be brought about. Like in a coil, in this case a phase shift of between voltage and current takes place so that the current is in advance of the voltage.

Fig. 7.12. shows vector diagram and line diagram illustrating these correlations.

Ideal capacitors have no effective resistance and no inductance.

Fig. 7.12. Current and voltage curves for a capacitive reactance Impedance Z

When effective resistances and reactances are connected together, either in series or parallel or series-parallel, then the equivalent resistance of the overall circuit is called impedance Z. It is the quotient of effective voltage value and effective current value.

Z = U/I

(7.9.)

where:

 Z impedance U effective voltage value I effective current value

[Z] = V/A = W

The reciprocal value of the impedance is called, admittance Y:

Y = 1/Z

Since the phase shift between current and voltage is 0° in effective resistances, 90° in reactances, the following important facts are found:

· An impedance causes a phase shift between current and voltage which is greater than 0° but smaller than 90°

The effective resistances and reactances resulting in the impedance are made up at right angles. This means that, in series connection, Z is always greater than the larger partial resistance but smaller than the algebraic sum of the two. Analogous conditions apply to the admittances in parallel connection.

Written in formular form, we have:

 series connection parallel connection (7.10.) (7.11.) tan j (u) = X/R (7.12.) tan j (i) = R/X (7.13.)

If, besides effective resistances, inductive and capacitive reactances are contained in a circuit, attention must be paid to the fact that the phase-shifting effect of the coil (I lags) behind U) is opposed to the action of a capacitor (I is in advance of U). Consequently, these two effects are partly neutralised or, in a special case, fully neutralised. The latter case is called resonance.

In a series connection of L and C, we have for X

 X = |XL - XC| If XL > XC, then X is inductive XL < XC, then X is capacitive XL = XC, then X = 0 (resonance)

In case of resonance, the highest current is flowing; it is only limited, by the effective resistance (current increase).

In a parallel connection of L and C, we have for X

1/X = |1/XL - 1/XC|

 X = |1/(1/XL - 1/XC)| If XL > XC then X is capacitive XL < XC then X is inductive XL = XC then X ® ¥ (resonance)

In case of resonance, the smallest current is flowing, namely, only the current through an effective resistance connected in parallel. When current is supplied at a constant rate, a maximum voltage drop is brought about (voltage increase).

Example 7.3.

A coil with an inductance L = 200 mH is connected in series with an effective resistance R = 100 W. A current of 500 mA with a frequency of 50 Hz is flowing through the circuit (Fig. 7.13.).

Fig. 7.13 Circuit for example 7.3

Draw the vector diagram for current and voltage true to scale. Calculate the partial voltages, the total voltage and the phase angle between current and voltage!

Given:

L = 0.2 H
R = 100 W
I = 0.5 A
f = 50 Hz

To be found:

vector diagram
UR; UL: U
j (u)

Solution:

UR = R I = 100 W · 0.5 A
UR = 50 V

UL = XLI
XL = wL = 2p f L = 2 · 3.14 · 50 1/s ··0.2 H
XL = 62.8 W
U1 = 63.8 W · 0.5 A
UL = 31.4 V

Now, the vector diagram can be drawn with the data obtained. The total voltage U 60 V and the phase angle j = 32° are indicated by the length of the vectors.

Fig. 7.14. Vector diagram for example 7.5.

Calculation:

U = Z · I

Z = 118 W
U = 118 W · 0.5 A
U = 59 V

Proof:

tan j = XL/R
tan j = 62.8/100 = 0.628
j = 32°

Example 7.4.

A capacitor with a capacity of C = 5 nF is connected, in parallel to an effective resistance R = 100 kW. A voltage of 10 V having a frequency of 300 Hz is applied to the circuit (Fig. 7.15.).

Fig. 7.15. Circuit for example 7.4.

Draw the vector diagram for current and voltage true to scale. Calculate the partial currents, the total current and phase angle between current and voltage!

Given:

C = 5 nF
R = 100 kW
U = 10 V
f = 300 Hz

To be found:

vector diagram
IR; IC; I
j (i)

Solution:

IR = U/R = 10 V/100 KW
IR = 100 µA

IC = U/XC
XC = 1/(wC) = 1/(2pfC) = 1/(2 · 3.14 · 300 1/s · 5 · 10-9 F)
XC = 106 KW
IC = 10 V/106 kW
IC = 94 µA

Now, the vector diagram can be drawn on the basis of the values obtained above.

The total current I » 140 µA and the phase angle of j 45° are indicated by the lengths of the vectors.

Calculation:

I = 137 µA

Proof:

I = U/Z

I = 10 V/73 kW = 137 µA
tan j = R/Xc
tan j = 100/106 = 0.943
j = 43 °

Example 7.5. Vector diagram for example 7.4.

An alternating voltage of 500 mV is applied to a series connection of R = 250 W, L = 200 µH and C = 125 pF. Calculate the maximum possible current and the frequency fr at which this current will flow!

Given:

R = 250 W
L = 200 µH
C = 125 pF
U = 500 mV

To be found:

Imax
fr (resonance frequency)

Solution:

Imax = U/R (case of resonance)
Imax = 0.5 V/250 W = 2 mA
Imax = 2 mA

resonance at XL = XC

XL = 2pfL
XC = 1/(2pfC)
2 · rfL = 1/(2pfrC)

Inversion for results in

fr2 = 1/(22 p2 LC)

fr » 1 MHz

There are effective resistance, reactances and impedances in the form of loads.

Effective resistance convert the electrical energy completely into heat energy; They are independent of frequency and do not cause phase shifts between current and voltage.

Storage elements such as coils and capacitors have a reactance. It is frequency dependent and causes a 90 ° phase shift between current and voltage. There are inductive and capacitive reactances.

In inductive reactances, the current lags behind the voltage, and in capacitive reactances, the current is in advance of the voltage.

Impedances are interconnections of effective resistances and reactances. They are dependent on frequency because of the reactance included in the system. The magnitude of the impedance can be found by diagrams or by calculation by geometric addition. Depending on the preponderance of the inductive component or the capacitive component, either the voltage is ahead of the current or vice versa. The phase angle is always between 0° and 90°. If, in one circuit, inductive and capacitive components are present, they neutralise each other partly or completely. The special case where the inductive reactance is equal to the capacitive reactance is called resonance. The frequency in the presence of which resonance occurs is called resonant frequency or frequency of resonance. When resonance is present, a circuit has the behaviour of an effective resistance.

Questions and problems:

1. What is the essence of effective (or acktive) resistances, reactances and impedances?

2. A coil has a reactance of 100 W at a frequency of 50 Hz. What is the size of the inductance?

3. Represent graphically the curve of the reactance in dependence of the frequency from 0 to 10 kHz for a coil having an inductance of 5 H.

4. At a frequency of 50 Hz, a capacitor has an impedance of about 65 W. What is the size of its capacity?

5. Represent graphically the curve of the impedance in dependence of the frequency from 0 to 10 kHz for a capacitor with a capacity of 100 µF!

6. A coil with a loss resistance (effective resistance) of 12 W connected in series has an impedance of Z = 20 W at a frequency of 50 Hz. The phase angle is 53°. Determine the magnitude of the inductive reactance and the inductance by graph and by calculation!

7. What is the frequency t which resonance occurs in a circuit?