CE00436-1 ELECTRICAL PRINCIPLES STEADY STATE ANALYSIS OF SINGLE PHASE CIRCUITS UNDER SINUSOIDAL...

CE00436-1 ELECTRICAL PRINCIPLES STEADY STATE ANALYSIS OF SINGLE PHASE CIRCUITS UNDER SINUSOIDAL EXCITATION

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Steady State response of Pure R,L and C & series & Parallel

circuits for sinusoidal excitation

http://www.apiit.edu.my/




Learning Outcomes

• How to compute the current and also voltage drops in the components, both in magnitude and phase, of the circuit?

• How to draw the complete phasor diagram, wave diagram showing the current and voltage drops relations?

• How to compute the total power and also power consumed in the components, along with power factor?

• How to compute the total Resistance, reactance and impedance of the R-L-C series circuit, fed from single phase ac supply of known frequency?

• How to compute the total conductance, susceptance and admittance of the R-L-C Parallel circuit, fed from single phase ac supply of known frequency?

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Purely Resistive circuit(R only)





Suppose a voltage v = Vp sin t is applied across a resistance R.

The resultant current i will be

tIR

tVRv

i PP

sinsin





Purely Inductive circuit(L only)





Pure Capacitive circuit(C only)





R-L series circuit





NOTE: Inductive loads have a lagging power factor & capacitive loads have a leading power factor





R-C series circuit





R-L-C series circuit





Summary of results of series AC circuit





parallel R-L circuit

Fig.(1) represents a parallel R-L circuit

excited by a sinusoidAt steady state,

(1)R

VI

R

and

(2)LL

V VI

jX j L

Apply Kirchhoff's current law,

R LI I I

Fig.(1)





1 1 (3)V VY

R j L

admittance ;

1where, conductance =

1inductive usceptance =

L

Y G jB

G mhoR

B s mhoX

1 1 1Here, Y = (4)

j

R j L R L

It may be noted that actual sign of B is ve





Also, the total current supplied by the source

in steady state lags the voltage by impedance

angle given by

1 1/tan

1/

L

R

1 = tan (5)

R

L

again, as , it can be further written asR Li i i 1

.v

i v dtR L

cos sinm mV V

t tR L

assuming cosmv V t 2 2

1 1 cos (6)mi V t

R L





parallel R-C circuitFig.(2) represents a parallel R-C circuit at

steady state and excited by sinusoidal voltage

source sinmv V t

; R

VHere I

R

1/CC

V VI j

X j C

1CX C

However, vectorially,

R CI I I

Fig.(1)





which gives,

1 1

1/I V

R j C

1or (1)I V j C V G jB

R

1

where, conductance = G mhoR

capacitive usceptance = + B s C mho

1and tan 1/

C

R

1 = tan (2)RC

Thus, it is evident that the current leads the voltage

by an angle given by expression (2)





also, ,R Ci i i it can be written as

v dvi CR dt

sin cosmm

Vt CV t

R

assuming sinmv V t

2

21Then sin (3)mi C V t

R

where is given by equation (2)





parallel RLC circuit

Fig. (1) below represents a steady state parallel RLC

circuit being energized by a voltage sinusoid

sinmv V tHere, R L Ci i i i

1.

v dvv dt C

R L dt

sin cos cos (1)m mm

V Vi t t CV t

R L

Fig.(1) Parallel RLC circuit





Let sin( )i A t

sin cos cos sin (2)A t A t

Equating the coefficient of sin and cos

in (1) and (2), we will get

t t

cosmV AR

and 1

sinmC V AL

Then1

tan (3)1/

CL

R





Obviously, the sign of the phase angle

1depends on the relative values of and C

L

In this context it may be noted that the inductive branch current is at 90 lagging the supply voltage while the LI

capacitive branch current is at 90 leading the supply CI

voltage. With proper selection of L and C, these twocurrents can mutually cancel each other and the net currentcan be resistive only. However, if the capacitive current CIis predominant, , the net current would be capacitive and iif the inductive current is predominant, the net current LI iwould be inductive




CE00436-1 ELECTRICAL PRINCIPLES STEADY STATE ANALYSIS OF SINGLE PHASE CIRCUITS UNDER SINUSOIDAL...

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