16 Single-phase parallela.c. circuits

At the end of this chapter you should be able to:

ž calculate unknown currents, impedances and circuit phaseangle from phasor diagrams for (a)R–L (b) R–C (c) L–C(d) LR–C parallel a.c. circuits

ž state the condition for parallel resonance in anLR–C circuit

ž derive the resonant frequency equation for anLR–C parallela.c. circuit

ž determine the current and dynamic resistance at resonance inanLR–C parallel circuit

ž understand and calculate Q-factor in anLR–C parallel circuit

ž understand how power factor may be improved

16.1 Introduction In parallel circuits, such as those shown in Figures 16.1 and 16.2, thevoltage is common to each branch of the network and is thus taken asthe reference phasor when drawing phasor diagrams.For any parallel a.c. circuit:

True or active power,P D VI cos� watts (W)

or P D IR2R watts

Apparentpower,S D VI voltamperes (VA)

Reactive power,Q D VI sin� reactive voltamperes (var)

Power factorD true power

apparentpowerD P

SD cos�

(These formulae are the same as for series a.c. circuits as used inChapter 15.)

Figure 16.1

16.2 R–L parallel a.c.circuit

In the two branch parallel circuit containing resistanceR and inductanceLshown in Figure 16.1, the current flowing in the resistance,IR, is in-phasewith the supply voltageV and the current flowing in the inductance,IL,lagsthe supply voltage by 90°. The supply currentI is the phasor sum ofIR and IL and thus the currentI lags the applied voltageV by an angle

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lying between 0° and90° (dependingon the values ofIR andIL), shownas angle� in the phasor diagram.

From the phasor diagram:

I D√I2

R C I2L�, (by Pythagoras’ theorem)

where IR D V

RandIL D V

XL

tan� D ILIR

, sin� D ILI

and cos� D IRI

(by trigonometric ratios�

Circuit impedance,Z D V

I

Problem1. A 20� resistor is connected in parallel with an induc-tance of 2.387 mH across a 60 V, 1 kHz supply. Calculate (a) thecurrent in each branch, (b) the supply current, (c) the circuit phaseangle, (d) the circuit impedance, and (e) the power consumed.

The circuit and phasor diagrams are as shown in Figure 16.1.

(a) Current flowing in the resistor IR D V

RD 60

20D 3 A

Currentflowing in the inductanceIL D V

XLD V

2�fL

D 60

2�1000�2.387ð 10�3�

D 4 A

(b) From the phasor diagram, supply current,I D√IR2 C I2

L�

D√32 C 42�

D 5 A

(c) Circuit phase angle,� D arctanILIR

D arctan(

4

3

)D 53.13°

D 53°8′ lagging

(d) Circuit impedance,Z D V

ID 60

5D 12Z

(e) Power consumedP D VI cos� D 60�5�cos 53°80� D 180 W

(Alternatively, power consumedP D IR2R D 3�220� D 180 W�

Further problems on R–L parallel a.c. circuits may be found inSection 16.8, problems 1 and 2, page 256.

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16.3 R–C parallel a.c.circuit

In the two branch parallel circuit containing resistanceR and capacitanceC shown in Figure 16.2,IR is in-phase with the supply voltageV and thecurrent flowing in the capacitor,IC, leadsV by 90°. The supply currentI is the phasor sum ofIR andIC andthus the currentI leads the appliedvoltageV by an angle lying between 0° and90° (dependingon the valuesof IR andIC), shown as angle in the phasor diagram.

From the phasor diagram:

I D√I2

R C I2C�, (by Pythagoras’ theorem)

where IR D V

RandIC D V

XC

tan˛ D ICIR

, sin˛ D ICI

andcos˛ D IRI

(by trigonometric ratios)

Circuit impedanceZ D V

IFigure 16.2

Problem2. A 30 µF capacitor is connected in parallel with an80 � resistor across a 240 V, 50 Hz supply. Calculate (a) thecurrent in each branch, (b) the supply current, (c) the circuit phaseangle, (d) the circuit impedance, (e) the power dissipated, and(f) the apparent power.

The circuit and phasor diagrams are as shown in Figure 16.2.

(a) Current in resistor,IR D V

RD 240

80D 3 A

Currentin capacitor,IC D V

XCD V(

1

2�fC

)

D 2�fCV

D 2�50�30ð 106�240�

D 2.262 A

(b) Supply current,I D√IR2 C IC2� D

√32 C 2.2622�

D 3.757 A

(c) Circuit phase angle, D arctanICIR

D arctan(

2.262

3

)

D 37°1′ leading

(d) Circuit impedance,Z D V

ID 240

3.757D 63.88Z

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(e) True or active power dissipated,P D VI cos˛

D 2403.757� cos 37°10

D 720 W

(Alternatively, true powerP D IR2 R D 3�280� D 720 W)

(f) Apparent power,S D VI D 240�3.757� D 901.7 VA

Problem3. A capacitor C is connected in parallel with a resistorR across a 120 V, 200 Hz supply. The supply current is 2 A at apower factor of 0.6 leading. Determine the values of C and R.

The circuit diagram is shown in Figure 16.3(a).

Power factorD cos� D 0.6 leading, hence � D arccos 0.6 D 53.13°

leading.

From the phasor diagram shown in Figure 16.3(b),

IR D I cos53.13° D 2�0.6�

D 1.2 A

and IC D I sin53.13° D 2�0.8�

D 1.6 A

(Alternatively, IR and IC can be measured from the scaled phasordiagram.)

Figure 16.3From the circuit diagram,

IR D V

Rfrom which R D V

IRD 120

1.2D 100Z

andIC D V

XCD 2�fCV, from which,C D IC

2�fV

D 1.6

2�200�120�

D 10.61 mF

Further problems onR–C parallel a.c. circuits may be found inSection 16.8, problems 3 and 4, page 256.

16.4 L –C parallel a.c.circuit

In the two branch parallel circuit containing inductance L and capacitanceC shown in Figure 16.4,IL lagsV by 90° andIC leadsV by 90°.

Theoreticallythere are three phasor diagrams possible — each depend-ing on the relative values ofIL andIC:

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Figure 16.4

(i) IL > IC (giving a supply current,I D IL � IC laggingV by 90°)

(ii) IC > IL (giving a supply current,I D IC � IL leadingV by 90°)

(iii) IL D IC (giving a supply current, I = 0).

The latter condition is not possible in practice due to circuit resistanceinevitably being present (as in the circuit described in Section 16.5).

For theL–C parallel circuit,IL D V

XL, IC D V

XC

I D phasordifference betweenIL andIC, and Z D V

I

Problem4. A pure inductance of 120 mH is connected in parallelwith a 25µF capacitor and the network is connected to a 100 V,50 Hz supply. Determine (a) the branch currents, (b) the supplycurrent and its phase angle, (c) the circuit impedance, and (d) thepower consumed.

The circuit and phasor diagrams are as shown in Figure 16.4.

(a) Inductive reactance,XL D 2�fL D 2�50�120ð 10�3�

D 37.70 �

Capacitivereactance,XC D 1

2�fCD 1

2�50�25ð 10�6�

D 127.3 �

Currentflowing in inductance,IL D V

XLD 100

37.70D 2.653 A

Current flowing in capacitor, IC D V

XCD 100

127.3D 0.786 A

(b) IL andIC areanti-phase. Hence supply current,

I D IL � IC D 2.653� 0.786D 1.867 A and the current lags thesupply voltageV by 90° (seeFigure 16.4(i))

(c) Circuit impedance,Z D V

ID 100

1.867D 53.56Z

(d) Power consumed,P D VI cos� D 100�1.867�cos 90°�

D 0 W

Problem5. Repeat Problem 4 for the condition when thefrequency is changed to 150 Hz.

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(a) Inductive reactance,XL D 2�150�120ð 10�3� D 113.1 �

Capacitivereactance,XC D 1

2�150�25ð 10�6�D 42.44 �

Currentflowing in inductance,IL D V

XLD 100

113.1D 0.884 A

Current flowing in capacitor,IC D V

XCD 100

42.44D 2.356 A

(b) Supply current,I D IC � IL D 2.356� 0.884D 1.472 A leading Vby 90° (seeFigure 4(ii))

(c) Circuit impedance,Z D V

ID 100

1.472D 67.93Z

(d) Power consumed,P D VI cos� D 0 W (since� D 90°)

From Problems 4 and 5:

(i) WhenXL < XC thenIL > IC andI lagsV by 90°

(ii) WhenXL > XC thenIL < IC andI leadsV by 90°

(iii) In a parallel circuit containing no resistance the power consumedis zero

Further problems on L–C parallel a.c. circuits may be found inSection 16.8, problems 5 and 6, page 256.

16.5 LR –C parallel a.c.circuit

In the two branch circuit containing capacitanceC in parallel with induc-tanceL and resistanceR in series (such as a coil) shown in Figure 16.5(a),the phasor diagram for the LR branch alone is shown in Figure 16.5(b)and the phasor diagram for theC branch is shown alone in Figure 16.5(c).Rotating each and superimposing on one another gives the completephasor diagram shown in Figure 16.5(d).

The currentILR of Figure 16.5(d) may be resolved into horizontal andvertical components. The horizontal component, shown as op isILR cos�1

and the vertical component, shown aspq is ILR sin�1. There are threepossible conditions for this circuit:

(i) IC > ILR sin�1 (giving a supply currentI leading V by angle�— as shown in Figure 16.5(e))

(ii) ILR sin�1 > IC (giving I lagging V by angle �— as shown inFigure 16.5(f))

(iii) IC D ILR sin�1 (this is called parallel resonance, see Section 16.6).

There are two methods of finding the phasor sum of currentsILR andIC in Figures 16.5(e) and (f). These are: (i) by a scaled phasor diagram,

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Figure 16.5

or (ii) by resolving each current into their ‘in-phase’ (i.e. horizontal) and‘quadrature’ (i.e. vertical) components,as demonstrated in problems 6and 7. With reference to the phasor diagrams of Figure 16.5:

Impedance of LR branch,ZLR D√R2 C XL

2�

Current, ILR D V

ZLRandIC D V

XC

SupplycurrentI D phasor sum ofILR andIC (by drawing)

D√

fILR cos�1�2 C ILR sin�1 ¾ IC�2g (by calculation)

where¾ means ‘the difference between’.

Circuit impedanceZ D V

I

tan�1 D VL

VRD XL

R, sin�1 D XL

ZLRand cos�1 D R

ZLR

tan� D ILR sin�1 ¾ ICILR cos�1

and cos� D ILR cos�1

I

Problem6. A coil of inductance 159.2 mH and resistance 40� isconnected in parallel with a 30µF capacitor across a 240 V, 50 Hzsupply. Calculate (a) the current in the coil and its phase angle,(b) the current in the capacitor and its phase angle, (c) the supplycurrent and its phase angle,(d) the circuit impedance, (e) the powerconsumed, (f) the apparent power, and (g) the reactive power. Drawthe phasor diagram.

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Figure 16.6

The circuit diagram is shown in Figure 16.6(a).

(a) For the coil, inductive reactanceXL D 2�fL

D 2�50�159.2 ð 10�3�

D 50 �

ImpedanceZ1 D√R2 C X2

L� D√402 C 502� D 64.03 �

Currentin coil, ILR D V

Z1D 240

64.03D 3.748 A

Branch phase angle�1 D arctanXL

RD arctan

(50

40

)D arctan1.25

D 51.34° D 51°20′ lagging

(seephasor diagram in Figure 16.6(b))

(b) Capacitive reactance,XC D 1

2�fCD 1

2�50�30ð 10�6�

D 106.1 �

Currentin capacitor,IC D V

XCD 240

106.1

= 2.262 A leading the supplyvoltage by 90°

(seephasor diagram of Figure 16.6(b)).

(c) The supply current I is the phasor sum ofILR and IC This may beobtained by drawing the phasor diagram to scale and measuring thecurrentI and its phase angle relative toV. (CurrentI will alwaysbe the diagonal of the parallelogram formed as in Figure 16.6(b)).

Alternatively the currentILR andIC maybe resolved into their hori-zontal (or ‘in-phase’) and vertical (or ‘quadrant’) components. Thehorizontal component ofILR is

ILR cos51°200� D 3.748cos 51°200 D 2.342 A

The horizontal component ofIC is IC cos90° D 0

Thusthe total horizontal component,IH D 2.342 A

The vertical component ofILR D �ILR sin51°200�

D �3.748sin 51°200

D �2.926 A

The vertical component ofIC D IC sin90°

D 2.262sin 90° D 2.262 A

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Thusthe total vertical component,IV D �2.926C 2.262

D −0.664 A

IH andIV areshown in Figure 16.7, from which,

I D√

[2.342�2 C �0.664�2] D 2.434 A

Angle � D arctan(

0.664

2.342

)D 15.83° D 15°500 lagging

Figure 16.7Hence the supply current I = 2.434 A lagging V by 15°50′.

(d) Circuit impedance,Z D V

ID 240

2.434D 98.60Z

(e) Power consumed,P D VI cos� D 240�2.434� cos 15°500

D 562 W

(Alternatively,P D IR2R D ILR2R (in this case)

D 3.748�240� D 562 W�

(f) Apparent power,S D VI D 240�2.434� D 584.2 VA

(g) Reactive power,Q D VI sin� D 240�2.434�sin 15°500�

D 159.4 var

Problem7. A coil of inductance 0.12 H and resistance 3 k� isconnected in parallel with a 0.02µF capacitor and is supplied at40 V at a frequency of 5 kHz. Determine (a) the current in the coil,and (b) the current in the capacitor. (c) Draw to scale the phasordiagram and measure the supply current and its phase angle; checkthe answer by calculation. Determine (d) the circuit impedance and(e) the power consumed.

The circuit diagram is shown in Figure 16.8(a).

(a) Inductive reactance,XL D 2�fL D 2�5000�0.12� D 3770�

Impedanceof coil, Z1 D√R2 C XL

2� D√

[3000�2 C 3770�2]

D 4818�

Currentin coil, ILR D V

Z1D 40

4818D 8.30 mA

Branch phase angle� D arctanXL

RD arctan

3770

3000

D 51.5° laggingFigure 16.8

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(b) Capacitive reactance,XC D 1

2�fCD 1

2�5000�0.02ð 10�6�

D 1592�

Capacitorcurrent,IC D V

XCD 40

1592

D 25.13 mA leading V by 90°

(c) Currents ILR and IC are shown in the phasor diagram ofFigure 16.8(b). The parallelogram is completed as shown and thesupply current is given by the diagonal of the parallelogram. ThecurrentI is measured as19.3 mA leading voltageV by 74.5°

By calculation,I D√

[ILR cos51.5°�2 C IC � ILR sin51.5°�2]

D 19.34 mA

and� D arctan(IC � ILR sin51.5°

ILR cos51.5°

)D 74.50°

(d) Circuit impedance,Z D V

ID 40

19.34ð 10�3D 2.068 kZ

(e) Power consumed,P D VI cos� D 40�19.34ð 10�3�cos74.50°�

D 206.7 mW

(Alternatively,P D IR2R D ILR2R D 8.30ð 10�3�23000�

D 206.7 mW)

Further problems on theLR–C parallel a.c. circuit may be found inSection 16.8, problems 7 and 8, page 256.

16.6 Parallel resonanceand Q-factor

Parallel resonance

Resonanceoccurs in the two branch network containing capacitanceC inparallel with inductanceL and resistanceR in series (see Figure 16.5(a))when the quadrature (i.e. vertical) component of currentILR is equal toIC. At this condition the supply currentI is in-phase with the supplyvoltageV.

Resonant frequency

When the quadrature component ofILR is equal toIC then:IC D ILR sin�1

(seeFigure 16.9)

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Figure 16.9

HenceV

XCD(

V

ZLR

)(XL

ZLR

), (from Section 16.5)

from which,ZLR2 D XCXL D 2�frL�

(1

2�frC

)D L

C(16.1)

Hence[√R2 C XL

2�]2 D L

CandR2 C XL

2 D L

C

Thus 2�frL�2 D L

C� R2 and2�frL D

√(L

C� R2

)

and fr D 1

2�L

√(L

C� R2

)D 1

2�

√√√√(

L

L2C� R2

L2

)

i.e. parallel resonant frequency,fr =1

2p

√√√√(

1LC

− R2

L2

)Hz

(WhenR is negligible, thenfr D 1

2�pLC�

, which is the same as for

series resonance.)

Current at resonance

Current at resonance,Ir D ILR cos�1 (from Figure 16.9)

D(

V

ZLR

)(R

ZLR

)(from Section 16.5)

D VR

Z2LR

Howeverfrom equation (16.1),Z2LR D L

C

henceI r =VRLC

=VRC

L(16.2)

The current is at aminimum at resonance.

Dynamic resistance

Since the current at resonance is in-phase with the voltage the impedanceof the circuit acts as a resistance. This resistance is known as thedynamicresistance,RD (or sometimes, the dynamic impedance).

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From equation (16.2), impedance at resonanceD V

IrD V(

VRC

L

) D L

RC

i.e. dynamic resistance,RD =L

RCohms

Rejector circuit

The parallel resonant circuit is often described as arejector circuit sinceit presents its maximum impedance at the resonant frequency and theresultant current is a minimum.

Q-factor

Currents higher than the supply current can circulate within the parallelbranches of a parallel resonant circuit, the current leaving the capacitorand establishing the magnetic field of the inductor, this then collapsing andrecharging the capacitor, and so on. TheQ-factor of a parallel resonantcircuit is the ratio of the current circulating in the parallel branches of thecircuit to the supply current, i.e. the current magnification.

Q-factor at resonanceD current magnificationD circulating current

supplycurrent

D ICIr

D ILR sin�1

Ir

D ILR sin�1

ILR cos�1D sin�1

cos�1

D tan�1 D XL

R

i.e. Q-factor at resonance=2pfr L

R

(which is the same as for a series circuit)

Note that in aparallel circuit the Q-factor is a measure ofcurrentmagnification, whereas in aseries circuit it is a measure ofvoltagemagnification.

At mains frequencies the Q-factor of a parallel circuit is usually low,typically less than 10, but in radio-frequency circuits the Q-factor can bevery high.

Problem8. A pure inductance of 150 mH is connected in parallelwith a 40µF capacitor across a 50 V, variable frequency supply.Determine (a) the resonant frequency of the circuit and (b) thecurrent circulating in the capacitor and inductance at resonance.

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Figure 16.10

The circuit diagram is shown in Figure 16.10.

(a) Parallel resonant frequency,fr D 1

2�

√(1

LC� R2

L2

)

However,resistanceR D 0. Hence,

fr D 1

2�

√(1

LC

)D 1

2�

√[1

150ð 10�3�40ð 10�6�

]

D 1

2�

√(107

15�4�

)

D 103

2�

√(1

6

)D 64.97 Hz

(b) Current circulating inL andC at resonance,

ICIRC D V

XCD V(

1

2�frC

) D 2�frCV

HenceICIRC D 2�64.97�40ð 10�6�50� D 0.816 A

Alternatively, ICIRC D V

XLD V

2�frLD 50

2�64.97�0.15�

D 0.817 A�

Problem9. A coil of inductance 0.20 H and resistance 60� isconnected in parallel with a 20µF capacitor across a 20 V, vari-able frequency supply. Calculate (a) the resonant frequency, (b) thedynamic resistance, (c) the current at resonance and (d) the circuitQ-factor at resonance.

(a) Parallel resonant frequency,

fr D 1

2�

√(1

LC� R2

L2

)

D 1

2�

√(1

0.20�20ð 10�6�� 60�2

0.2�2

)

D 1

2�

p250000� 90 000�

D 1

2�

p160000� D 1

2�400�

D 63.66 Hz

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(b) Dynamic resistance,RD D L

RCD 0.20

60�20ð 10�6�D 166.7 Z

(c) Current at resonance,Ir D V

RDD 20

166.7D 0.12 A

(d) Circuit Q-factor at resonanceD 2�frL

RD 2�63.66�0.2�

60D 1.33

Alternatively, Q-factor at resonanceD current magnification (for aparallel circuit)D Ic/Ir

Ic D V

XcD V(

1

2�frC

) D 2�frCV D 2�63.66�20ð 10�6�20�

D 0.16 A

HenceQ-factorD IcIr

D 0.16

0.12D 1.33, as obtained above

Problem10. A coil of inductance 100 mH and resistance 800�is connected in parallel with a variable capacitor across a 12 V,5 kHz supply. Determine for the condition when the supply currentis a minimum: (a) the capacitance of the capacitor, (b) the dynamicresistance, (c) the supply current, and (d) the Q-factor.

(a) The supply current is a minimum when the parallel circuit is atresonance.

Resonant frequency,fr D 1

2�

√(1

LC� R2

L2

)

Transposing for C gives:2�fr�2 D 1

LC� R2

L2

2�fr�2 C R2

L2D 1

LC

C D 1

L

{2�fr�2 C R2

L2

}

WhenL D 100 mH, R D 800� andfr D 5000Hz,

C D 1

100ð 10�3

{2�5000�2 C 8002

100ð 10�3�2

}

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D 1

0.1[�2108 C 0.64�108]F

D 106

0.110.51ð 108�µF D 0.009515mF or 9.515 nF

(b) Dynamic resistance,RD D L

CRD 100ð 10�3

9.515ð 10�9�800�

D 13.14 kZ

(c) Supply current at resonance,Ir D V

RDD 12

13.14ð 103D 0.913 mA

(d) Q-factor at resonance =2�frL

RD 2�5000�100ð 10�3�

800D 3.93

Alternatively,Q-factor at resonanceD Ic

IrD V/Xc

IrD 2�frCV

Ir

D 2�5000�9.515ð 10�9�12�

0.913ð 10�3

D 3.93

Further problems on parallel resonance and Q-factor may be found inSection 16.8, problems 9 to 12, page 257.

16.7 Power factorimprovement

For a particular power supplied, a high power factor reduces the currentflowing in a supply system and therefore reduces the cost of cables,switch-gear, transformers and generators. Supply authorities use tariffswhich encourage electricity consumers to operate at a reasonably highpower factor. Industrial loads such as a.c. motors are essentially induc-tive (R–L) and may have a low power factor. One method of improving(or correcting) the power factor of an inductive load is to connect astatic capacitorC in parallel with the load (see Figure 16.11(a)). Thesupply current is reduced fromILR to I, the phasor sum ofILR andIC, and the circuit power factor improves from cos�1 to cos�2 (seeFigure16.11(b)).

Problem11. A single-phase motor takes 50 A at a power factor of0.6 lagging from a 240 V, 50 Hz supply. Determine (a) the currenttaken by a capacitor connected in parallel with the motor to correctthe power factor to unity, and (b) the value of the supply currentafter power factor correction.

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Figure 16.11

The circuit diagram is shown in Figure 16.12(a).

(a) A power factor of 0.6 lagging means that cos� D 0.6i.e. � D arccos 0.6 D 53°80

HenceIM lagsV by 53°80 asshownin Figure 16.12(b).

If the power factor is to be improved to unity then the phase differ-ence between supply currentI and voltageV is 0°, i.e. I is in phasewith V as shown in Figure 16.12(c). For this to be so,IC mustequal the length ab, such that the phasor sum ofIM and IC is I.ab D IM sin53°80 D 500.8� D 40 A

Hencethe capacitor current I c must be40 A for the power factorto be unity.

(b) Supply currentI D IM cos53°80 D 500.6� D 30 A

Problem12. A motor has an output of 4.8 kW, an efficiency of80% and a power factor of 0.625 lagging when operated from a240 V, 50 Hz supply. It is required to improve the power factor to0.95 lagging by connecting a capacitor in parallel with the motor.Determine (a) the current taken by the motor, (b) the supply currentafter power factor correction, (c) the current taken by the capacitor,(d) the capacitance of the capacitor, and (e) the kvar rating of thecapacitor.

(a) Efficiency D power output

power inputhence

80

100D 4800

power input

Powerinput D 4800

0.8D 6000W

Hence,6000D VIM cos� D 240�IM�0.625�,

since cos� D p f. D 0.625

Thus current taken by the motor,IM D 6000

240�0.625�D 40 A

The circuit diagram is shown in Figure 16.13(a).

The phase angle betweenIM andV is given by:

� D arccos 0.625D 51.32° D 51°190, hencethe phasor diagram is asshown in Figure 16.13(b).

(b) When a capacitorC is connected in parallel with the motor a currentIC flows which leadsV by 90°. The phasor sum ofIM andIC givesthe supply current I, and has to be such as to change the circuitpower factor to 0.95 lagging, i.e. a phase angle of arccos 0.95 or18°120 lagging,asshown in Figure 16.13(c).

Figure 16.12

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Figure 16.13

The horizontal component ofIM (shownas oa)D IM cos51°190

D 40cos 51°190

D 25 A

The horizontal component ofI (also given by oa)D I cos 18°120

D 0.95 I

Equatingthe horizontal components gives: 25D 0.95 I

Hence the supply current after p.f. correction,I D 25

0.95D 26.32 A

(c) The vertical component ofIM (shownas ab)D IM sin51°190

D 40sin 51°190

D 31.22 A

The vertical component ofI (shown as ac) D I sin 18°120

D 26.32sin 18°120

D 8.22 A

The magnitude of the capacitor currentIC (shownas bc) is givenby ab � ac, i.e. 31.22� 8.22 D 23 A

(d) CurrentIC D V

XcD V(

1

2�fC

) D 2�fCV,

from which,C D IC2�fV

D 23

2�50�240�F D 305mF

(e) kvar rating of the capacitorD VIc1000

D 240�23�

1000D 5.52 kvar

In this problem the supply current has been reduced from 40 A to26.32 A without altering the current or power taken by the motor. Thismeans that the size of generating plant and the cross-sectional area ofconductors supplying both the factory and the motor can be less — withan obvious saving in cost.

Problem13. A 250 V, 50 Hz single-phase supply feeds thefollowing loads (i) incandescent lamps taking a current of 10 Aat unity power factor, (ii) fluorescent lamps taking 8 A at a powerfactor of 0.7 lagging, (iii) a 3 kVA motor operating at full load andat a power factor of 0.8 lagging and (iv) a static capacitor. Deter-mine, for the lamps and motor, (a) the total current, (b) the overallpower factor and (c) the total power. (d) Find the value of the staticcapacitor to improve the overall power factor to 0.975 lagging.

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Figure 16.14

A phasor diagram is constructed as shown in Figure 16.14(a), where8 A is lagging voltageV by arccos 0.7, i.e. 45.57°, and the motor currentis 3000/250, i.e. 12 A laggingV by arccos 0.8, i.e. 36.87°

(a) The horizontal component of the currents

D 10 cos 0° C 12cos 36.87° C 8cos 45.57°

D 10C 9.6 C 5.6 D 25.2 A

The vertical component of the currents

D 10 sin 0° � 12sin 36.87° � 8sin 45.57°

D 0 � 7.2 � 5.713D �12.91 A

From Figure 16.14(b), total current,IL D√

[25.2�2 C 12.91�2]

D 28.31 A

at a phase angle of� D arctan(

12.91

25.2

), i.e. 27.13° lagging

(b) Power factorD cos� D cos 27.13° D 0.890 lagging

(c) Total power,P D VIL cos� D 250�28.31�0.890� D 6.3 kW

(d) To improve the power factor, a capacitor is connected in parallelwith the loads. The capacitor takes a currentIC suchthat the supplycurrent falls from 28.31 A toI, lagging V by arccos 0.975, i.e.12.84°. The phasor diagram is shown in Figure 16.15.

Figure 16.15

oaD 28.31 cos 27.13° D I cos12.84°

HenceI D 28.31cos 27.13°

cos12.84°D 25.84 A

CurrentIC D bc D ab � ac�

D 28.31sin 27.13° � 25.84sin 12.84°

D 12.91� 5.742

D 7.168 A

Ic D V

XcD V(

1

2�fC

) D 2�fCV

HencecapacitanceC D Ic2�fV

D 7.168

2�50�250�F

= 91.27 mF

Thus to improve the power factor from 0.890 to 0.975 lagging a91.27µF capacitor is connected in parallel with the loads.

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16.8 Further problemson single-phase parallel

a.c. circuits

R–L parallel a.c. circuit

1 A 30 � resistor is connected in parallel with a pure inductance of3 mH across a 110 V, 2 kHz supply. Calculate (a) the current in eachbranch, (b) the circuit current, (c) the circuit phase angle, (d) thecircuit impedance, (e) the power consumed, and (f) the circuit powerfactor.

[(a) IR D 3.67 A, IL D 2.92 A (b) 4.69 A (c) 38°300lagging(d) 23.45� (e) 404 W (f) 0.783 lagging]

2 A 40 � resistance is connected in parallel with a coil of inductanceL and negligible resistance across a 200 V, 50 Hz supply and thesupply current is found to be 8 A. Draw a phasor diagram to scaleand determine the inductance of the coil. [102 mH]

R–C parallel a.c. circuit

3 A 1500 nF capacitor is connected in parallel with a 16� resistoracross a 10 V, 10 kHz supply. Calculate (a) the current in eachbranch, (b) the supply current, (c) the circuit phase angle, (d) thecircuit impedance, (e) the power consumed, (f) the apparent power,and (g) the circuit power factor. Draw the phasor diagram.

[(a) IR D 0.625 A, IC D 0.943 A (b) 1.13 A (c) 56°280 leading(d) 8.85 � (e) 6.25 W (f) 11.3 VA (g) 0.55 leading]

4 A capacitorC is connected in parallel with a resistanceR across a60 V, 100 Hz supply. The supply current is 0.6 A at a power factorof 0.8 leading. Calculate the value ofR andC.

[R D 125�, C D 9.55 µF]

L –C parallel a.c. circuit

5 An inductance of 80 mH is connected in parallel with a capacitanceof 10 µF across a 60 V, 100 Hz supply. Determine (a) the branchcurrents, (b) the supply current, (c) the circuit phase angle, (d) thecircuit impedance and (e) the power consumed.

[(a) IC D 0.377 A, IL D 1.194 A (b) 0.817 A(c) 90° lagging(d) 73.44 � (e) 0 W]

6 Repeat problem 5 for a supply frequency of 200 Hz.[(a) IC D 0.754 A, IL D 0.597 A (b) 0.157 A

(c) 90° leading(d) 382.2 � (e) 0 W]

LR –C parallel a.c. circuit

7 A coil of resistance 60� and inductance 318.4 mH is connectedin parallel with a 15µF capacitor across a 200 V, 50 Hz supply.

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Calculate(a) the current in the coil, (b) the current in the capacitor,(c) the supply current and its phase angle, (d) the circuit impedance,(e) the power consumed, (f) the apparent power and (g) the reactivepower. Draw the phasor diagram.

[(a) 1.715 A (b) 0.943 A (c) 1.028 A at 30°540 lagging(d) 194.6 � (e) 176.5 W (f) 205.6 VA (g) 105.6 var]

8 A 25 nF capacitor is connected in parallel with a coil of resis-tance 2 k� and inductance 0.20 H across a 100 V, 4 kHz supply.Determine (a) the current in the coil, (b) the current in the capac-itor, (c) the supply current and its phase angle (by drawing a phasordiagram to scale, and also by calculation), (d) the circuit impedance,and (e) the power consumed.

[(a) 18.48 mA (b) 62.83 mA (c) 46.17 mA at 81°290leading(d) 2.166 k� (e) 0.683 W]

Parallel resonance and Q-factor

9 A 0.15 µF capacitor and a pure inductance of 0.01 H are connected inparallel across a 10 V, variable frequency supply. Determine (a) theresonant frequency of the circuit, and (b) the current circulating inthe capacitor and inductance.

[(a) 4.11 kHz (b) 38.73 mA]

10 A 30 µF capacitor is connected in parallel with a coil of inductance50 mH and unknown resistanceR across a 120 V, 50 Hz supply. Ifthe circuit has an overall power factor of 1 find (a) the value ofR,(b) the current in the coil, and (c) the supply current.

[(a) 37.7 � (b) 2.94 A (c) 2.714 A]

11 A coil of resistance 25� and inductance 150 mH is connected inparallel with a 10µF capacitor across a 60 V, variable frequencysupply. Calculate (a) the resonant frequency, (b) the dynamic resis-tance, (c) the current at resonance and (d) the Q-factor at resonance.

[(a) 127.2 Hz (b) 600�(c) 0.10 A (d) 4.80]

12 A coil of resistance 1.5 k� and 0.25 H inductance is connected inparallel with a variable capacitance across a 10 V, 8 kHz supply.Calculate (a) the capacitance of the capacitor when the supply currentis a minimum, (b) the dynamic resistance, and (c) the supply current.

[(a) 1561 pF (b) 106.8 k� (c) 93.66µA]

Power factor improvement

13 A 415 V alternator is supplying a load of 55 kW at a power factorof 0.65 lagging. Calculate (a) the kVA loading and (b) the currenttaken from the alternator. (c) If the power factor is now raised tounity find the new kVA loading.

[(a) 84.6 kVA (b) 203.9 A (c) 84.6 kVA]

14 A single phase motor takes 30 A at a power factor of 0.65 laggingfrom a 240 V, 50 Hz supply. Determine (a) the current taken by the

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capacitorconnected in parallel to correct the power factor to unity,and (b) the value of the supply current after power factor correction.

[(a) 22.80 A (b) 19.5 A]

15 A motor has an output of 6 kW, an efficiency of 75% and a powerfactor of 0.64 lagging when operated from a 250 V, 60 Hz supply. Itis required to raise the power factor to 0.925 lagging by connecting acapacitor in parallel with the motor. Determine (a) the current takenby the motor, (b) the supply current after power factor correction,(c) the current taken by the capacitor, (d) the capacitance of thecapacitor and (e) the kvar rating of the capacitor.

[(a) 50 A (b) 34.59 A (c) 25.28 A (d) 268.2µF (e) 6.32 kvar]

16 A 200 V, 50 Hz single-phase supply feeds the following loads:(i) fluorescent lamps taking a current of 8 A at a power factor of0.9 leading, (ii) incandescent lamps taking a current of 6 A at unitypower factor, (iii) a motor taking a current of 12 A at a power factorof 0.65 lagging. Determine the total current taken from the supplyand the overall power factor. Find also the value of a static capacitorconnected in parallel with the loads to improve the overall powerfactor to 0.98 lagging. [21.74 A, 0.966 lagging, 21.68µF]

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