Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE) PDF Download

In the last lesson, two points were described:

1. How to solve for the impedance, and current in an ac circuit, consisting of single element, R / L / C?
2. How to solve for the impedance, and current in an ac circuit, consisting of two elements, R and L / C, in series, and then draw complete phasor diagram?

In this lesson, the solution of currents in simple circuits, consisting of resistance R, inductance L and/or capacitance C connected in series, fed from single phase ac supply, is presented. Then, the circuit with all above components in parallel is taken up. The process of drawing complete phasor diagram with current(s) and voltage drops in the different components is described. The computation of total power and also power consumed in the different components, along with power factor, is explained. One example of series circuit are presented in detail, while the example of parallel circuit will be taken up in the next lesson.

Keywords: Series and parallel circuits, impedance, admittance, power, power factor.

After going through this lesson, the students will be able to answer the following questions;

1. How to compute the total reactance and impedance / admittance, of the series and parallel circuits, fed from single phase ac supply?
2. How to compute the different currents and also voltage drops in the components, both in magnitude and phase, of the circuit?
3. How to draw the complete phasor diagram, showing the currents and voltage drops?
4. How to compute the total power and also power consumed in the different components, along with power factor?

Solution of Current in R-L-C Series Circuit 

Series (R-L-C) circuit

The voltage balance equation for the circuit with R, L and C in series (Fig. 15.1a), is

The current, i is of the form,

As described in the previous lesson (#14) on series (R-L) circuit, the current in steady state is sinusoidal in nature. The procedure given here, in brief, is followed to determine the form of current. If the expression for   is substituted in the voltage equation, the equation shown here is obtained, with the sides (LHS & RHS) interchanged.

The steps to be followed to find the magnitude and phase angle of the current I , are same as described there (#14).
So, the phase angle is   and the magnitude of the current is   

where the impedance of the series circuit is 

Alternatively, the steps to find the rms value of the current I, using complex form of impedance, are given here.

The impedance of the circuit is

(a) Inductive

In this case, the circuit is inductive, as total reactance (ωL − (1/ω C)) is positive, under the condition (ωL > (1/ωCL)). The current lags the voltage by φ (taken as positive), with the voltage phasor taken as reference. The power factor (lagging) is less than 1 (one), as 0° ≤ φ ≤ 900° . The complete phasor diagram, with the voltage drops across the components and input voltage (OA), and also current (OB ), is shown in Fig. 15.1b. The voltage phasor is taken as reference, in all cases. It may be observed that 

using the Kirchoff’s second law relating to the voltage in a closed loop. The phasor diagram can also be drawn with the current phasor as reference, as will be shown in the example given here. The expression for the average power is  . The power is only consumed in the resistance, R, but not in inductance/capacitance (L/C), in all three cases.

In this case, the circuit is inductive, as total reactance is positive, under the condition  The current lags the voltage by φ (positive). The power factor (lagging) is less than 1 (one), . The complete phasor diagram, with the voltage drops across the components and input voltage (OA), and also current (OB), is shown in Fig. 15.1b. The voltage phasor is taken as reference, in all cases. It may be observed that

using the Kirchoff’s second law relating to the voltage in a closed loop. The phasor diagram can also be drawn with the current phasor as reference, as will be shown in the example given here. The expression for the average power is  . The power is only consumed in the resistance, R, but not in inductance/capacitance (L/C), in all three cases.

(b) Capacitive 

The circuit is now capacitive, as total reactance  is negative, under the condition . The current leads the voltage by φ , which is negative as per convention described in the previous lesson. The voltage phasor is taken as reference here. The complete phasor diagram, with the voltage drops across the components and input voltage, and also current, is shown in Fig. 15.1c. The power factor (leading) is less than 1 (one), as  being negative. The expression for the average power remains same as above.

The third case is resistive, as total reactance  is zero (0), under the condition  . The impedance is  . The current is now at unity power factor i.e. the current and the voltage are in phase. The complete phasor diagram, with the voltage drops across the components and input (supply) voltage, and also current, is shown in Fig. 15.1d. This condition can be termed as ‘resonance’ in the series circuit, which is described in detail in lesson #17. The magnitude of the impedance in the circuit is minimum under this condition, with the magnitude of the current being maximum. One more point to be noted here is that the voltage drops in the inductance, L and also in the capacitance, C, is much larger in magnitude than the supply voltage, which is same as the voltage drop in the resistance, R. The phasor diagram has been drawn approximately to scale.

It may be observed here that two cases of series (R-L & R-C) circuits, as discussed in the previous lesson, are obtained in the following way. The first one (inductive) is that of (a), with C very large, i.e. 1/ ωC ≈ 0 , which means that C is not there. The second one (capacitive) is that of (b), with L not being there (L or ωL = 0).

Example 15.1

A resistance, R is connected in series with an iron-cored choke coil (r in series with L). The circuit (Fig. 15.2a) draws a current of 5 A at 240 V, 50 Hz. The voltages across the resistance and the coil are 120 V and 200 V respectively. Calculate,
(a) the resistance, reactance and impedance of the coil,
(b) the power absorbed by the coil, and
(c) the power factor (pf) of the input current

I (OB) = 5 A    VS (OA) = 240 V      f = 50 Hz    ω= 2π f
The voltage drop across the resistance V₁ (OC) = I . R = 120 V

The resistance, R = V₁/ I = 120/ 5 = 24 Ω

The voltage drop across the coil V₂(CA) = I. Z_L = 200 V

The impedance of the coil, 
From the phasor diagram (Fig. 15.2b),



The power factor (pf) of the input current 
The phase angle of the total impedance, 
Input voltage, 
The total impedance of the circuit, 

The total resistance of the circuit, 
The resistance of the coil, 
The reactance of the coil, 
The inductance of the coil, 

The phase angle of the coil,

The power factor (pf) of the coil, 
The copper loss in the coil 

Example 15.2

An inductive coil, having resistance of 8 Ω and inductance of 80 mH, is connected in series with a capacitance of 100 μF across 150 V, 50 Hz supply (Fig. 15.3a). Calculate, (a) the current, (b) the power factor, and (c) the voltages drops in the coil and capacitance respectively.

Solution

The impedance of the coil, 
The total impedance of the circuit,

The current drawn from the supply,

The current is, 
The power factor (pf) 

Please note that the current phasor is taken as reference in the phasor diagram (Fig. 15.3b) and also here. The voltage drop in the coil is,

The voltage drop in the capacitance in, 

Solution of Current in Parallel Circuit 

Parallel circuit

The circuit with all three elements, R, L & C connected in parallel (Fig. 15.4a), is fed to the ac supply. The current from the supply can be computed by various methods, of which two are described here.

First method

The current in three branches are first computed and the total current drawn from the supply is the phasor sum of all three branch currents, by using Kirchoff’s first law related to the currents at the node. The voltage phasor (V) is taken as reference.

All currents, i.e. three branch currents and total current, in steady state, are sinusoidal in nature, as the input (supply voltage is sinusoidal of the form, 

Three branch currents are obtained by the procedure given in brief.

Total (supply) current, i is

The two equations given here are obtained by expanding the trigonometric form appearing in the last term on RHS, into components of cosω t and sinω t , and then equating the components of cos ωt and sin ωt from the last term and last but one (previous) .

From these equations, the magnitude and phase angle of the total (supply) current are,




where, the magnitude of the term (admittance of the circuit) is,

Please note that the admittance, which is reciprocal of impedance, is a complex quantity. The angle of admittance or impedance, is same as the phase angle, φ of the current I , with the input (supply) voltage taken as reference phasor, as given earlier.

Alternatively, the steps required to find the rms values of three branch currents and the total (suuply) current, using complex form of impedance, are given here. Three branch currents are

Of the three branches, the first one consists of resistance only, the current, I_R is in phase with the voltage (V). In the second branch, the current, I_L lags the voltage by 90° , as there is inductance only, while in the third one having capacitance only, the current, I_C leads the voltage 90°. All these cases have been presented in the previous lesson. 

The total current is

where,

The two cases are as described earlier in series circuit.

(a) Inductive

In this case, the circuit being inductive, the current lags the voltage by φ (positive), as . This condition is in contrast to that derived in the case of series circuit earlier. The power factor is less than 1 (one). The complete phasor diagram, with the three branch currents along with total current, and also the voltage, is shown in Fig. 15.4b. The voltage phasor is taken as reference in all cases. It may be observed there that

The Kirchoff’s first law related to the currents at the node is applied, as stated above. The expression for the average power is 

The power is only consumed in the resistance, R, but not in inductance/capacitance (L/C), in all three cases. 

(b) Capacitive

The circuit is capacitive, as  The current leads the voltage by φ (φ being negative), with the power factor less than 1 (one). The complete phasor diagram, with the three branch currents along with total current, and also the voltage, is shown in Fig. 15.4c.

The third case is resistive, as  This is the same condition, as obtained in the case of series circuit. It may be noted that two currents, I_L and I_C , are equal in magnitude as shown, but opposite in sign (phase difference being 180°), and the sum of these currents   The total current is in phase with the voltage  the power factor being unity. The complete phasor diagram, with the three branch currents along with total current, and also the voltage, is shown in Fig. 15.4d. This condition can be termed as ‘resonance’ in the parallel circuit, which is described in detail in lesson #17. The magnitude of the impedance in the circuit is maximum (i.e., the magnitude of the admittance is minimum) under this condition, with the magnitude of the total (supply) current being minimum.

The circuit with two elements, say R & L, can be solved, or derived with C being large 

Second method

Before going into the details of this method, the term, Admittance must be explained. In the case of two resistance connected in series, the equivalent resistance is the sum of two resistances, the resistance being scalar (positive). If two impedances are connected in series, the equivalent impedance is the sum of two impedances, all impedances being complex. Please note that the two terms, real and imaginary, of two impedances and also the equivalent one, may be positive or negative. This was explained in lesson no. 12.

If two resistances are connected in parallel, the inverse of the equivalent resistance is the sum of the inverse of the two resistances. If two impedances are connected in parallel, the inverse of the equivalent impedance is the sum of the inverse of the two impedances. The inverse or reciprocal of the impedance is termed ‘Admittance’, which is complex. Mathematically, this is expressed as

As admittance (Y) is complex, its real and imaginary parts are called conductance (G) and susceptance (B) respectively. So, Y = G + jB . If impedance,  with X being positive, then the admittance is 

Please note the way in which the result of the division of two complex quantities is obtained. Both the numerator and the denominator are multiplied by the complex conjugate of the denominator, so as to make the denominator a real quantity. This has also been explained in lesson no. 12.

The magnitude and phase angle of Z and Y are

To obtain the current in the circuit (Fig. 15.4a), the steps are given here. The admittances of the three branches are


The total admittance, obtained by the phasor sum of the three branch admittances, is

The total impedance of the circuit is

The total current in the circuit is obtained as

where the magnitude of current is 

The current is the same as obtained earlier, with the value of Y substituted in the above equation. This is best illustrated with an example, which is described in the next lesson. The solution of the current in the series-parallel circuits will also be discussed there, along with some examples.