A.C. CIRCUITS We know that the current, whose magnitude and direction changes

12.1. INTRODUCTION

We know that the current, whose magnitude and direction changes continuously, is called an alternating current. The circuit through which A.C. is passed is called an a.c. circuit. The study of a.c. circuits is also known as a.c. theory.

In d.c. circuits the flow of current is opposed by resistance only and they obey Ohm's law. But in a.c. circuits the reactance (inductive or capacitive) also opposes the flow of current in addition to the resistance. hence, the Opposition to the flow of current does not depend on resistance only and a.c. circuits do not obey Ohm's law. The total opposition offered by an a.c. circuit to the flow of A.G. is called its impedance. Its symbol is Z and its unit is ohm.

If R is replaced by Z in Ohm's law R= V/I then the resulting formula Z = V/I can be used in a.c. circuit's calculations.

12.2.PURE RESISTIVE A.C. CIRCUIT

A pure resistive circuit has no reactance and the voltage remains in phase with the current. For a d.c. circuit the calculation of power is done by using the formula P = V . I or P = P . R or P = V2/R. Since there is no reactance in a pure resistive a.c. circuit, hence Z = R therefore, the calculation of P can be done by any of the above stated formula but only if the *effective or R.M.S. value of voltage and current have been used.

12.3.SERIES R-L CIRCUIT

The voltage VL leads the current IL by 90° in a pure inductive circuit. But, if the resistance of the circuit is also considered or an external resistance is added to the circuit, then the circuit becomes a series R-L circuit. The angle of lag of the circuit current is then found to be less than 90°, see Fig. 12.1. In a series circuit

Fig. 12.1. Series R-L circuit

*The effective vlaue of A.C. is 0.707 times of its maximum value and it is the value which produces the same amount of heat as produced with D.C.

The angle of lag flout the impedance triangle OAB

Example 12.1. if the mistime of a 10 henrys choke is 50 ohms then calculate (i) circuit I impedance. (ii) power factor, (iii) power consumption at 250 volts 50 Hertz A.C.

Solution. Given :            Voltage V = 250 volts

                                     Frequency. f = 50 hertz

                                     Inductance, 1, 10 henrys

                                     Resistance, R = 50 ohms

                    

            

12.4. SERIES R-C CIRCUIT

The current lc leads the voltage V_C by 90° in a pure capacitive circuit. But if the resistance of the circuit is also considered or an external resistance is added to the circuit, then the circuit becomes a series R-C circuit. The angle of lead of current is found to he less than 90°, see Fig. 12.2. In a series R-C circuit,

 

Fig. 12.2. Series R.C. circuit

Example 12.2. A 10 ohms resistor is connected in series with a 50 micro farad capacitor. If the source voltage is 250 volts, 50 Hz then calculate (i) impedance; (ii) power consumption.

Solution. Given :             Capacitance, C = 50 µF

                                                               = 50 x 10^-6 F

                                          Resistance, R = 10 ohms

                                              Voltage, V = 250 volts

                                           Frequency, f = 50 hertz

12.5. SERIES L-C CIRCUIT

The inductor voltage VL leads the current I in a series L-C- circuit while the capacitor voltage V_C lags behind the current I. In this way, V_L and V_C act in the directors exactly opposite to each other, hence there may be following three possibilities :

(i) When V_L > V_C then V = V_L — V_C and it will lead the current I by 90° and Z = X_L — X_C, see Fig. 12.3 (b).

(ii) When V_L < V_C then V = V_C — V_L and it will lag behind the current / by 90° and Z =X_C – X_L , see Fig. 12.3 (c).

Fig. 12.3. Series L-C circuit 

(iii) When V_L = V_C then V = 0 and Z = 0 and current I will become infinite.

The condition is termed as resonance.

Note. Since, all circuits have little or more resistance present in them, hence, the current cannot reach infinity but it will reach its maximum value.

12.6. SERIES L-C-R CIRCUIT

In a series L-C-R circuit, inductor voltage VL leads the current and the capacitor voltage V_C lags behind the current. In this way, V_L and V_C act in opposite directions. The resultant circuit voltage (by ∆ OAB)

                                   V² = V_R² + ( V_L – V_c)²

Fig. 12.4. Series L-C-R circuit

Hence. in resonance condition (X_L = X_C), the circuit reactance becomes zero, the circuit is remained a Pure resistive circuit and the voltage and current become in-phase.

Example 12.3. The inductance of a circuit is 0.25 henrys. If the value of capacitor is 50 micro rarads, the value of resistor is 10 ohms and the source voltage is 100 volts 50 hertz, calculate Impedance, (ii) current, (iii) power factor.

Solution. Given : Inductance, L = 0.25 H

                            Capacitance, C = 50 µF = 50 X 10^-6 F

                                   Voltage, V = 100 V

                                Frequency, F = 50 Hz

                               Resistance, R = 10Ω

              Inductive reactance, X_L = 2 π.f.L

                                                        = 2 x 3.14 x 50 x 0.25

                                                       = 78.5 Ω

12.7. PARALLEL R-L CIRCUIT

In a parallel R-L circuit, the resistor current (I_R) is in-phase with the voltage and the inductor current (I_L) lags behind the voltage by 90°. The angle of lag of the resultant current (I) remains less than 90⁰. In parallel circuit,

Fig. 12.5 Parallel R-L circuit 

Example 12.4. A circuit comprises of a 15 ohms resistor connected across a 50 milli henrys choke. If the source voltage is 100 volts, 50 hertz then calculate (i) impedance, (ii) power factor.

Solution. Given :         Inductance, L = 50 mH = 50 x 10^-3 H

                                       Resistance, R = 15Ω

                                        Frequency, f = 50 Hz

•                   Inductive reactance, X_L = 2 π.f.L

                                                               = 2 x 3.14 x 50 x 50 x 10^-3

                                                               = 15.7Ω

Fig. 12.7. Parallel L-C circuit

(iii) When 1 t= I then the resultanct current will be zero (I = 0) and

Hence, the magnitude of current will become zero and the magnitude of impedance will become infinite. The condition is termed as resonance.

Note. Since all circuits have little or more resistance present in them, therefore, the current cannot be zero and the impedance cannot be infinite. Of course, it can be maximum.

12.11. PARALLEL L-C-R CIRCUIT

In a parallel L-C-R circuit, the inductor current (I_L) acts in a direction quite opposite to that of capacitor current (I_C), hence, the resultant current :

Fig. 12.8. Parallel L-C-R circuit

Example 12.6. If a 10 ohms resistor. a 25 milli henrys choke and a 10 micro farad capacitor are connected across 100 volts, 50 hertz A.C. then calculate (i) current in each branch, (ii) total current, (iii) impedance

Solution, Given,                          Source voltage, F – 100 V

                                                                  Frequency, f – 5 Hz

                                                                 Resistance, R – 10 Ω

                                                                 Inductance, L – 25mH

                                                               Capacitance, C – 10 µF

Therefore,                              Inductive reactance, X_L – 2 π.f.L

                                                                                  = 2 x 3.14 x 50 x 25 x10^-3

                                                                                = 7.85 Ω