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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

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

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

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

The current, i is of the form,

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

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  Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE) is substituted in the voltage equation, the equation shown here is obtained, with the sides (LHS & RHS) interchanged.

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

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  Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE) and the magnitude of the current is  Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE) 

where the impedance of the series circuit is  Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)

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

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

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

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 Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE) . The power is only consumed in the resistance, R, but not in inductance/capacitance (L/C), in all three cases.

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

In this case, the circuit is inductive, as total reactanceSolution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE) is positive, under the condition Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE) The current lags the voltage by φ (positive). The power factor (lagging) is less than 1 (one), Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE). 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

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

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 Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE) . The power is only consumed in the resistance, R, but not in inductance/capacitance (L/C), in all three cases.

(b) Capacitive 

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

The circuit is now capacitive, as total reactance Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE) is negative, under the condition Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE). 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 Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE) being negative. The expression for the average power remains same as above.

The third case is resistive, as total reactance Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE) is zero (0), under the condition Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE) . The impedance is Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE) . The current is now at unity power factorSolution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE) 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.

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

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

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

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. ZL = 200 V

The impedance of the coil, Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
From the phasor diagram (Fig. 15.2b),
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
The power factor (pf) of the input current  Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
The phase angle of the total impedance, Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
Input voltage,  Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
The total impedance of the circuit,  Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
The total resistance of the circuit,  Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
The resistance of the coil,  Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
The reactance of the coil,  Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
The inductance of the coil,  Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)

The phase angle of the coil,

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

The power factor (pf) of the coil, Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
The copper loss in the coil Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)

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

Solution

Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
The impedance of the coil,  Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
The total impedance of the circuit,

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

The current drawn from the supply,

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

The current is,  Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
The power factor (pf)  Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)

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,

Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
The voltage drop in the capacitance in,  Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)

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.

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

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, 

Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
Three branch currents are obtained by the procedure given in brief.
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)

Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
Total (supply) current, i is
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)

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) .

Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
From these equations, the magnitude and phase angle of the total (supply) current are,
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
where, the magnitude of the term (admittance of the circuit) is,
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)

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

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

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

The total current is

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

The two cases are as described earlier in series circuit.

(a) Inductive

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

In this case, the circuit being inductive, the current lags the voltage by φ (positive), as Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE). 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

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

The Kirchoff’s first law related to the currents at the node is applied, as stated above. The expression for the average power is Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)

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 Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE) 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 Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE) This is the same condition, as obtained in the case of series circuit. It may be noted that two currents, IL and IC , are equal in magnitude as shown, but opposite in sign (phase difference being 180°), and the sum of these currents  Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE) The total current is in phase with the voltage Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE) 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.

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

The circuit with two elements, say R & L, can be solved, or derived with C being large Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)

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

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

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

As admittance (Y) is complex, its real and imaginary parts are called conductance (G) and susceptance (B) respectively. So, Y = G + jB . If impedance, Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE) with X being positive, then the admittance is 

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

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

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

Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
To obtain the current in the circuit (Fig. 15.4a), the steps are given here. The admittances of the three branches are
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
The total admittance, obtained by the phasor sum of the three branch admittances, is
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)

Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
The total impedance of the circuit is
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
The total current in the circuit is obtained as
Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)
where the magnitude of current is  Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE)

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.

The document Solution of Current in AC Series & Parallel Circuits | Basic Electrical Technology - Electrical Engineering (EE) is a part of the Electrical Engineering (EE) Course Basic Electrical Technology.
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FAQs on Solution of Current in AC Series & Parallel Circuits - Basic Electrical Technology - Electrical Engineering (EE)

1. What is the difference between series and parallel circuits?
Ans. In a series circuit, the components are connected in a single path, and the current remains the same throughout the circuit. In a parallel circuit, the components are connected in multiple paths, and the current divides among the branches.
2. How does current behave in an AC series circuit?
Ans. In an AC series circuit, the current varies with time. It reaches its maximum value (peak) and then decreases to zero, changes direction, and reaches the maximum value in the opposite direction. This pattern repeats continuously.
3. How can I calculate the total current in a parallel AC circuit?
Ans. To calculate the total current in a parallel AC circuit, you need to find the sum of the individual branch currents. The total current is equal to the algebraic sum of the branch currents, taking into account their directions.
4. Can I mix AC and DC in a series or parallel circuit?
Ans. It is possible to mix AC and DC components in a series or parallel circuit. However, it is important to consider the characteristics of each type of current and ensure that the components are properly connected and can handle both AC and DC.
5. How does impedance affect the current in an AC series circuit?
Ans. Impedance, which includes resistance and reactance, affects the current in an AC series circuit. The higher the impedance, the lower the current flow. Impedance is the opposition to the flow of AC current and depends on the frequency and components in the circuit.
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