Solution of Current in AC Parallel and Series-parallel Circuits

 

In the last lesson, the following points were described:

 

1. How to compute the total impedance/admittance in series/parallel circuits?

 

2. How to solve for the current(s) in series/parallel circuits, fed from single phase ac supply, and then draw complete phasor diagram?

 

3. How to find the power consumed in the circuit and also the different components, and the power factor (lag/lead)?

 

In this lesson, the computation of impedance/admittance in parallel and series-parallel circuits, fed from single phase ac supply, is presented. Then, the currents, both in magnitude and phase, are calculated. The process of drawing complete phasor diagram is described. The computation of total power and also power consumed in the different components, along with power factor, is explained. Some examples, of both parallel and series-parallel circuits, are presented in detail.

Keywords: Parallel and series-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 impedance/admittance, of the parallel and series-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?

This lesson starts with two examples of parallel circuits fed from single phase ac supply. The first example is presented in detail. The students are advised to study the two cases of parallel circuits given in the previous lesson.

Example 16.1

The circuit, having two impedances of Ω+=)158(1jZ and Ω−=)86(2jZ in parallel, is connected to a single phase ac supply (Fig. 16.1a), and the current drawn is 10 A. Find each branch current, both in magnitude and phase, and also the supply voltage.

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Solution

Ω°∠=+=∠93.6117)158(11jZφ Ω°−∠=−=−∠13.5310)86(22jZφ

AjOCI)010(010)(0+=°∠=°∠

The admittances, using impedances in rectangular form, are,

1322222210)0.800.60(1008686868611−−Ω⋅+=+=++=−=−∠=∠jjjjZYφφ

Alternatively, using impedances in polar form, the admittances are,

13222210)0.800.60(13.531.013.530.1011−−Ω⋅+=°∠=°−∠=−∠=∠jZYφφ

The total admittance is,

332110)1.2868.87(10)]0.800.60()9.5168.27[(−−⋅+=⋅++−=+=∠jjjYYYφ

1377.171007.92−−Ω°∠⋅=

The total impedance is,

Ω−=°−∠=°∠⋅=∠=−∠−)315.3343.10(77.1786.1077.171007.92113jYZφφ

The supply voltage is

VZIVVAB°−∠=°−∠×=−∠⋅°∠=−∠77.176.10877.17)86.1010(0)(φφ

Vj)15.3343.103(−=

The branch currents are,

AZVODI°−∠=°+°−∠⎟⎠⎞⎜⎝⎛=∠−∠=−∠7.7939.6)93.6177.17(0.176.108)(1111φφθ B I=10AFig. 16.1 (a) Circuit diagramA Z2 = (6 - j8) Ω Z1 = (8 + j15) Ω I1 I2

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Aj)286.6143.1(−=

2211()0()(10.00.0)(1.1436.286)(8.8576.286)10.8635.36IOEIIOCODOCCEjjjAθθ∠=∠°−∠−−=−=+−−=+=∠°

Alternatively, the current is, 2I

2222108.6()(17.7753.13)10.8635.3610.0VIOEAZφθφ∠−⎛⎞∠==∠−°+°=∠⎜⎟∠−⎝⎠

Aj)285.6857.8(+=

The phasor diagram with the total (input) current as reference is shown in Fig. 16.1b.

Alternative Method

Ω°∠=+=−++=−∠+∠=′∠′565.2665.15)714()86()158(2211jjjZZZφφφ

Ω−=°−∠=°−°−°∠⎟⎠⎞⎜⎝⎛×=′∠′−∠⋅∠=+⋅=−∠)315.3343.10(77.1786.10)565.2613.5393.61(65.150.100.1722112121jZZZZZZZZφφφφ

The supply voltage is

()(1010.86)17.77108.617.77(103.4333.15)ABVVIZjVφ∠−=⋅=×∠−°=∠−°=−

The branch currents are,

2111210.010.0()(53.1326.565)6.3979.715.65(1.1426.286)×⎛⎞∠−==∠−°−°=∠−°⎜⎟+⎝⎠=−ZIODIAZZjAθ Fig. 16.1 (b) Phasor diagram53.13° I2= 10.8661.90 108.63 V Ф = 17.8° I1 = 6.4 A D θ1 = 79.7° θ1 = 35.35 O E VABC I = 10A

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221()()(10.00.0)(1.1426.286)(8.8586.286)10.86235.36IOEIIOCODOCCEjjjAAθ∠=−−=−=+−−=+=∠°

Alternatively, the current is, 2I

1221210.017.0()(61.9326.565)10.8635..3615.65(8.8586.286)ZIOEIAZZjAθ×⎛⎞∠==∠°−°=∠⎜⎟+⎝⎠=+

Example 16.2

The power consumed in the inductive load (Fig. 16.2a) is 2.5 kW at 0.71 lagging power factor (pf). The input voltage is 230 V, 50 Hz. Find the value of the capacitor C, such that the resultant power factor of the input current is 0.866 lagging.

Solution

WKWP2500105.25.23=⋅== V = 230 V = 50 Hz f

The power factor in the inductive branch is )(71.0coslagL=φ

The phase angle is °≈°==−4577.44)71.0(cos1Lφ

()2500cos230cos=⋅=⋅=LLLLIIVPφφ

AVPILL31.1571.02302500cos=×==φ

AIAILLLL87.1045sin32.15sin;87.1071.031.15cos=°×==×=φφ

The current is, LIAjILL)87.1087.10(4531.15−=°−∠=−∠φ

The power consumed in the circuit remains same, as the capacitor does not consume any power, but the reactive power in the circuit changes. The active component of the total current remains same as computed earlier.

AIILL87.10coscos==φφ

The power factor of the current is cos)(866.0

The phase angle is °==−30)866.0(cos1φ

The magnitude of the current is AI55.12866.0/87.10==

The current is AjI)276.687.10(3055.12−=°−∠=−∠φ + - C Fig. 16.2 (a) Circuit diagramI ILIC230 V L O A D

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The current in the capacitor is

AjjjIIILLC°∠==−−−=−∠−−∠=°∠90504.4504.4)87.1087.10()276.687.10(90φφ

This current is the difference of two reactive currents,

AIIILLC504.487.10276.6sinsin−=−=−=−φφ

The reactance of the capacitor, C is Ω====066.51504.423021CCIVCfXπ

The capacitor, C is FXfCCμππ33.621033.62066.515021216=⋅=××==−

The phasor diagram with the input voltage as reference is shown in Fig. 16.2b.

Example 16.3

An inductive load (R in series with L) is connected in parallel with a capacitance C of 12.5 Fμ (Fig. 16.3a). The input voltage to the circuit is 100 V at 31.8 Hz. The phase angle between the two branch currents, (LII=1) and (CII=2) is , and the current in the first branch is . Find the total current, and also the values of R & L. °120AIIL5.01== 230 VC 15.3 A IC4.5 A φ45° Fig. 16.2 (b) Phasor diagram B IC A B C = 12.5 μF I2I R D L + - 100 V Fig. 16.3 (a) Circuit diagram 1=0.5A I

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Solution

f = 31.8 Hz sradf/2008.3122≈×==ππω V = 100 V

C = AI5.01=FF6105.125.12−⋅=μ

Ω=⋅×==−400)105.12200/(1)/(16CXCω

The current in the branch no. 2 is

AjXjVIC)25.00.0(9025.090)400/100(90400/0100)/(902+=°∠=°∠=°−∠°∠=−=°∠

The current in the branch no. 1 is 1115.0φφ−∠=−∠I

The phase angle between and is 1I2I°=+°120901φ

So, °=°−°=30901201φ

AjI)25.0433.0(305.0301−=°−∠=°−∠

The impedance of the branch no. 1 is,

Ω+=°∠=°∠=°−∠°∠=+=∠)0.1002.173(3020030)5.0/100(30/0)(111jIVXjRZLφ

Ω=2.173R Ω==0.100LXLω

So, mHHXLL500105005.0200/100/3=⋅====−ω

The total current is,

AjjjIII°∠=+=+−=°∠+°−∠=°∠0433.0)0.0433.0(25.0)25.0433.0(9030021

The total impedance is,

Ω+=°∠=°∠=°∠°∠=+′=°∠)0.00.231(00.2310)433.0/100(0/0)0(0jIVjRZ

The current, I is in phase with the input voltage, V .

The total admittance is )90/1()30/1(90021211°−∠+°∠=°∠+−∠=°∠ZZYYYφ

The total impedance is )9030/()9030(02121°−∠+°∠°−∠⋅°∠=°∠ZZZZZ

Any of the above values can be easily calculated, and then checked with those obtained earlier. The phasor diagram is drawn in Fig. 16.3b.

Solution of Current in Series-parallel Circuit

Series-parallel circuit

The circuit, with a branch having impedance , in series with two parallel branches having impedances, and , shown in Fig. , , is connected to a single phase ac supply. 1Z2Z3Z

The impedance of the branch, AB is 11φφ∠=∠ZZABAB

333322221;1φφφφ∠=−∠∠=−∠ZYZY

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The admittance of the parallel branch, BC is

3322332211φφφφφ∠+∠=−∠+−∠=−∠ZZYYYBCBC

The impedance of the parallel branch, BC is

()(32323322332233221φφφφφφφφφφ+∠∠+∠=∠⋅∠∠+∠=−∠=∠ZZZZZZZZYZBCBCBCBC

The total impedance of the circuit is

BCBCBCBCABABACACZZZZZφφφφφ∠+∠=∠+∠=∠11

The supply current is

ACACACZVIφφ∠°∠=−∠0

The current in the impedance is 2Z

()33223342φφφφφ∠+∠∠−∠=∠ZZZIIAC

Thus, the currents, along with the voltage drops, in all branches are calculated. The phasor diagram cannot be drawn for this case now. This is best illustrated with the following examples, where the complete phasor diagram will also be drawn in each case.

Example 16.4

Find the input voltage at 50 Hz to be applied to the circuit shown in Fig. 16.4a, such that the current in the capacitor is 8 A? 30°BAI = 0.4330.25 I20.5 I1Z1DI2 = 0.25 Fig. 16.3 (b) Phasor diagram86.6 V100 V 50 V I R1 = 5 Ω L1 = 25.5 mHR3 = 7 ΩL3 = 38.2 mHBA R2 = 8 Ω318 μFI12I=8AFig. 16.4 (a) Circuit diagram ••••I1D

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Solution

f = 50 Hz sradf/16.3145022=×==ππω

HL0255.01= HL0382.03= FFC610318318−⋅==μ

Ω=×==80255.016.31411LXω Ω=×==120382.016.31433LXω

Ω=⋅×==−10)1031816.314/(1)/(162CXωAjI)08(0802+=°∠=°∠

Ω°∠=+=+=∠58434.9)85(1111jXjRZφ

Ω°−∠=−=−=−∠34.51806.12)108(2222jXjRZφ

Ω°∠=+=+=∠74.5989.13)127(3333jXjRZφ

VjVZIVAC)8064(34.5145.10234.51)806.120.8(02222−=°−∠=°−∠×=−∠⋅°∠=−∠φφ

AjAZVIAC)25.106.3(34.10986.10)5834.51(434.945.10211211−−=°−∠=°+°−∠⎟⎠⎞⎜⎝⎛=∠−∠=−∠φφθ

AjjjIII°−∠=−=+++−=°∠+−∠=−∠77.66154.11)25.104.4()0.00.8()25.106.3(02113θθ

VjVZIVCBCB)96.18764.153(03.793.154)74.5977.66()89.13154.11(333−=°−∠=°+°−∠×=∠⋅−∠=−∠φθθ

VjjjVVVCBCBACABAB°−∠=−=−+−=−∠+−∠=−∠44.242.239)96.09764.217()96.18764.153()0.800.64(2θφθ

The phasor diagram with the branch current, as reference, is shown in Fig. 16.4b. 2I I1 = 10.86A 58° 51.34° 66.77°11.15AI I2 = 8A VDB = 154.9V VAB = 239.2Fig. 16.4 (b) Phasor diagram7.03°VAD = 102.45 V

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Example 16.5

A resistor of 50 in parallel with an inductor of 30 mH, is connected in series with a capacitor, C (Fig. 16.5a). A voltage of 220 V, 50 Hz is applied to the circuit. Find, Ω

(a) the value of C to give unity power factor,

(b) the total current, and

(c) the current in the inductor

Solution

= 50 Hz fsradf/16.3145022=×==ππω

Ω=50R V = 220 V HmHL03.01030303=⋅==−

Ω=×==24.9403.016.314LXLω

The admittance, is, ADY

1395.2702264.010)61.100.20(24.94150111−−Ω°−∠=⋅−=+=+=−∠jjXjRYlADADφ

The impedance, ADZ is,

Ω+=°∠=°−∠=∠=∠)7.2002.39(95.2717.44)95.2702264.0/(1)/(1jYZADADADADφφ

The impedance of the branch (DB) is )/(1[CjXjZCDBω−=−=.

As the total current is at unity power factor (upf), the total impedance, is resistive only. ABZ

)7.20(02.399000CDBADADABABXjZZjRZ−+=°−∠+∠=+=°∠φ

Equating the imaginary part, Ω==7.20)/(1CXcω

The value of the capacitance C is, A Fig. 16.5 (a) Circuit diagram DB L = 30 mHR = 50 Ω 230 V 50 Hz C ••

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FXCCμω8.153108.1537.2016.314116=⋅=×==−

So, Ω°∠=+=+=°∠002.39)0.002.39(00jjRZABAB

The total current is,

AjZVIAB°∠=+=°∠=°∠°∠=°∠064.5)0.064.5(0)02.39/0.220(0/00

The voltage, is, ADV

VjVZIVADADADAD)73.1160.220(95.2705.24995.27)17.4464.5(0+=°∠=°∠×=∠⋅°∠=∠φφ

The current in the inductor, is, LI

AjAXVILADADLL)335.224.1(05.6264.2)9095.27()24.94/05.249(90/−=°−∠=°−°∠=°∠∠=∠φθ

The phasor diagram is shown in Fig. 16.5b.

Example 16.6

In the circuit (Fig. 16.6a) the wattmeter reads 960 W and the ammeter reads 6 A. Calculate the values of nd . LCCSIIIVV,,,, aCX 28° 5.64 62° IL = 2.64 A IR = 4.98 28°A D B 116.7 V 116.7 V 249.0 220 V Fig. 16.5 (b) Phasor diagram(i) (ii) jxL = j 8 Ω VCICC ICA R2 = 6 Ω ILE A + - D B R1 = 10 Ω WI Fig. 16.6 (a) Circuit diagram • • ••

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Solution

In this circuit, the power is consumed in two resistance, and only, but not consumed in inductance L, and capacitance C. These two components affect only the reactive power. 1R2R

P = 960 W I = 6 A Ω=101R

Ω=62RΩ=8LX

Total power is, WIIRIRIPLLL9606360610)6(2222212=⋅+=⋅+×=⋅+⋅=

or, WIL60036096062=−=⋅

So, AIL106/600==

The impedance of the inductive branch is,

Ω°∠=+=+=13.5310)86(2jXjRZLL

The magnitude of the voltage in the inductive branch is,

VZIVVLLCDB1001010=×=⋅==

Assuming as reference, the current, is, °∠=0100DBVLI

AjAZVILLDBLL)86(13.531013.53)10/100(/0−=°−∠=°−∠=∠°∠=−∠φφ

The current, °∠=°−∠°∠==°∠90)/(90/0 90CCDBCCXVXVIjI

The total current is )8(6)86(90−+=+−=°∠+−∠=∠CCCLLIjIjjIIIφφ

So, AIIC6)8()6(22=−+=

or, 36)6()8(3622==−+CI

So, AIC8=

The capacitive reactance is, Ω===5.128/100/CDBCIVX

The total current is AjI°∠=+=°∠06)06(0

or, it can be written as,

AjjjIIICLL°∠=+=+−=°∠+−∠=°∠06)06(8)86(900φ

The voltage is. ADV

VjRIjRIVAD)060(0600)106(0)()0(0011+=°∠=°∠×=°∠⋅=+⋅°∠=°∠

The voltage, is. ABSVV=

VjVVVVDBADABS)0160(00160)10060(000+=°∠=°∠+=°∠+°∠=°∠=

The current, I is in phase with ABSVV= , and also . DBV

The total impedance is,

Ω+=°∠=°∠=°∠°∠=°∠+°∠=°∠)0.067.26(067.260)6/160(0/0000jIVZZZABDBADAB

The impedance, is, DBZ

Ω+=°∠=°∠=°∠°∠=°∠−°∠=°∠)0.067.16(067.160)6/100(0/0000jIVZZZDBADABDB

Both the above impedances can be easily obtained using the circuit parameters by the method given earlier, and then checked with the above values. The impedance, can DBZ

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be obtained by the steps given in Example 16.3. The phasor diagram is shown in Fig. 16.6b.

Starting with the examples of parallel circuits, the solution of the current in the series-parallel circuit, along with the examples, was taken up in this lesson. The problem of resonance in series and parallel circuits will be discussed in the next lesson. This will complete the module of single phase ac circuits A B8 A 53.13°AI = 6A60 V8 A6 ADAB = 160 V (VO) Fig. 16.6 (b) Phasor diagram

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Problems

16.1 Find the impedance, Z_ab in the following circuits (Fig. 16.7a-b): (check with admittance diagrams in complex plane)

(a) (b)

16.2 A resistor (R) of 50 Ω in parallel with a capacitor (C) of 40 μF, is connected in series with a pure inductor (L) of 30 mH to a 100 V, 50 Hz supply. Calculate the total current and also the current in the capacitor. Draw the phasor diagram.

16.3 In a series-parallel circuit (Fig.16.8), the two parallel branches A and B, are in series with the branch C. The impedances in Ω are, Z_A = 5+j, Z_B = 6-j8, and Z_C = 10+j8. The voltage across the branch, C is (150+j0) V. Find the branch currents, I_A and I_B, and the phase angle between them. Find also the input voltage. Draw the phasor diagram.

16.4 A total current of 1A is drawn by the circuit (Fig.16.9) fed from an ac voltage, V of 50 Hz. Find the input voltage. Draw the phasor diagram. a • Q • • R P ZA = (5 + j6) Ω ZC = (10 + j8) Ω ZB = (6 - j8) Ω 150 V Fig. 16.8

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List of Figures

Fig. 16.1 (a) Circuit diagram (Ex. 16.1) (b) Phasor diagram

Fig. 16.2 (a) Circuit diagram (Ex. 16.2) (b) Phasor diagram

Fig. 16.3 (a) Circuit diagram (Ex. 16.3) (b) Phasor diagram

Fig. 16.4 (a) Circuit diagram (Ex. 16.4) (b) Phasor diagram

Fig. 16.5 (a) Circuit diagram (Ex. 16.5) (b) Phasor diagram

Fig. 16.6 (a) Circuit diagram (Ex. 16.6) (b) Phasor diagram

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