Basic Electrical and Electronics Engineering: Unit I: Electrical Circuits
RLC Series Circuit
Basic Circuit diagram, Phasor diagram, Formula with Solved Example Problems | Electrical Circuits
Apparent power, S = VI, Active power, P = VI cos ϕ, Reactive power, Q = VI sin ϕ
RLC
SERIES CIRCUIT
Fig
1.35 shows the RLC series circuit
Where
R
- Resistance in ohm
L
- Inductance in H
C
- Capacitance in F
V
- Input voltage
I
- Current through the circuit
VR
- Voltage drop across R = IR
VL
- Voltage drop across L = IXL
VC
- Voltage drop across C = IX.
Impedance
Phasor
diagram is shown in fig 1.36
Example: 17
A
resistor of 20Ω, inductor of 0.2 H and a capacitor of 150 µF are connected in
series and fed by a 230 V, 50 Hz, AC supply. Find (i) XL (ii) XC
(iii) Impedance (iv) Current (v) Power factor (vi) Apparent power (vii) Active
power (Viii) Reactive power.
XL
= 2πƒL = 2 × π × 50 × 0.2
=
62.83 Ω
=
21.22 Ω
=
46.16 Ω
Current
I = V/Z
=
230/46.16
=
4.98 A
Power
factor, cos ϕ = R/Z
=
20/46.16
=
0.433
Apparent
power, S
VI
= 230 × 4.98
=
1145.4 VA
Active
power, P
P
= VI cos ϕ
=
230 × 4.98 × 0.433
=
495.95 W
Reactive
power, Q
Q
= VI sin ϕ
=
230 × 4.98 × 0.901
=
1032 VAR
Basic Electrical and Electronics Engineering: Unit I: Electrical Circuits : Tag: : Basic Circuit diagram, Phasor diagram, Formula with Solved Example Problems | Electrical Circuits - RLC Series Circuit
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