Problems on RL, LC, RLC circuits for electronic circuits.pdf

Problems on RL, LC, RLC.pdf

1. RL, RC & RLC Circuits -
Problems

2. ㅣ
RL Circuit

P.7 A resistance of 150 is connected in series with an inductance of 0.25 Henry across
220 Volts, 50 Hz a.c. supply: Find the

Inductive reactance

Total impedance

iii. Current through the circuit

iv. Voltage across the resistance

v. Voltage across the inductance

vi. Power factor

vii. Phase angle between voltage and current
viii. Active, reactive and apparent power
ix. Time equation for voltage and current

. x. Also draw the phasor diagram.

Given

R=150
L=0.25 Henry
Vous = 220 Volts
f=50 Hz 220V, 50 Hz

The circuit arrangement is as shown in the Figure. 6.43 below re)

3. = _ ㄴㄴ

RL Circuit

i. Inductive reactance (X,) = 2afL

= 240% 50 x 0.25
=78.54 0

Total impendance |Z! = (RP +(X,)? = 79.960

| Vias 220 e
through the circuit (Lx) = 121 = 5996 7275 Amps

resistance (V,) = Teg XR

RMS

2.75 x 15 = 41.27 Volts
= Lys * (X)
= 2.75 x 70.54 = 216.09 Volts

ㆍ COUT)

4. ㆍ
RL Circuit

R 25

vi. Power factor (cos 0 ) = Zi = 7096 7 = 0.1875

IR
vii. Phase angle ‘6’ between voltage and current = cos] 5/= cos 1(0.1875) = 79°.18
ang 8 2

viii, Active power P = Vases Lag COS >

RMS "RMS

1

220 x 2.75 x 0,1875 = 113.49 Watts
Reactive power Q = Vus lanes SINO

= 220 x 2.75 x 0.9822 = 594.25 VARs.
Apparent power S = Vas Tag = 220 x 2:75

= 605 Volt amperes.

5. ' ——_

RL Circuit

ix. Time equation for voltage is given by
vo =V,sinot

where AY Vans” V3 Vases = VE * 220 = 311.12 volts

and © =2nf=2xxx50=314 rad/sec
bins » © Y =31112sin314t volts

Time equation for current is given by

i =I, sin(ot- 4)

V, 31112
=== = 3.89
Ta = 796 “9 Amps

i =3.89sin(314t-79”.18)

6. ㅣ
RC Circuit

P.8 A resistance of 202 is connected in series with an capacitance of 10,
115V, 50 Hz a.c supply. Find the

i Capacitive reactance

ii. Total impedance

iii. Current through the circuit.

iv. Voltage across the resistance and capacitance.
v. Power factor and phase angle

8.

Active, reactive and apparent power.
vii. Time equation for voltage and current.
viii. Also draw the phasor diagram.

Solution
Given R-C series circuit Ay =
- R=200
C=104F =10x10%F Yo
f=50 Hz
Vs = 115 V

115V, 50 Hz

7. ㅣ
RC Circuit

Figure 6.45

1
TT #09

Total impedance 121 = JR? + xz =319.020

115

= 319.02 = 0.3604 Amps

ss Resistance (Vz) = JawsxR
= 7.209 Volts
capt

8. ㅣ —
RC Circuit

Voltage across capacitance (Vc) u X)

= 114.73 volts

= 0.0626
v. Power factor cos 6 = iz) = 319.02 =
E

Phase angle $ = cos" (7) = cos \319.02) = 86 40
Reactive power (P) = Vous haus Sin ©

= 115 x 0.3604 x 0.9980

= 41.364 VARs
Apparent power (S) Mas Ml

1180; 3604

= 41.446 VAs

9. >
RC Circuit

vii. Time equation for voltage is given by
y =V,, sin (ot)
where 다, = V2 Varas = 162.63 volts
© = 2nf = 314 rad/sec

| = © v = 162.63 sin (314 t) volts
Time equation for current is given by
GE 1 =1,,,sin (ot +0)
Laa = V2 Tens = V2 x 0.3604 = 0.5098 Amps

i = 0.5098 sin (314t + 86°.40) Amps

mad

10. ' =

RLC Circuit

P.9 An R-L- C series circuit has a resistance of 15 Q, inductance 150 milli henry and
capacitance of 25 uF connected in series across 200 V, 50 Hz supply: Find the values

of inductive and capacitive reactances, impedance, current, power factor, power,
voltage across R, L and C. Also draw the phasor diagram.

Solution

Given, R-L- C circuit 1562 150 mH

2snF
R =15Q AMO HA
L = 150 milli Henry -ー マ ーーー
50 x 107 = 0.15H Me Yi 3
(a =25 pF
=25 x 10* Farads .
1 =50 Hz 200V . 50 Hz
Vaws =200 V Figure 6.47
2 X = 2nfL = 2x Rx 50x 015 = 47.1232
MA O
0 = Inf © 2xrx50x25x10* ~

121

[PAX Y

= ‚(15% + (47.123 - 127.32)? =81.590

11. RLC Circuit

R 15
Power factor (cos $) = Zi = 8159 = 0.1838

= (5):

Phase angle (9). = jz1) = 79.406

Sing = Sin(79°.406) = 0.9829

= Views _ as
121 7 8159

e ee

= 200 x 2.451 x 79.406

= 90.109 Watts

= Vans 566 Sin 6

= 200 x 2.451 x 0.9829

= 481.817 Vars. _

= 2.451 Amps

12. ㆍ
RLC Circuit

Apparent power (S) Vin lass

= 200 x 2.45

= 490.2 Volt amperes.
Voltage across resistance (V) = lps * R

= 2.451 x 15

36.765 Volts
Voltage across inductance (V,) = lays x (X,) = 2.451 x 47.123
312.06 Volts

Phasor Diagram (Here X。> X,)

9 E

7 leads v by 7 の 4 v lags 7 by 794
ㆍ Figure 6.48

13. SSS
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