Phasor Diagram for Series RLC Circuits

frequency f in Hz

50

resistance R in Ω

5

inductance L in mH

50

capacitance C in μF

250

This Demonstration shows a phasor diagram in an AC series RLC circuit. The circuit consists of a resistor with resistance

R

, an inductor with inductance

L

, and a capacitor with capacitance

C

. The current in an RLC series circuit is determined by the differential equation



I

=



E

0

R+j(

X

L

-

X

C

)

,

where

j=

-1

and

e(t)=

E

0

sin(ωt)

is the AC emf driving the circuit. The angular frequency

ω

is related to the frequency

f

in hertz (Hz) by

ω=2πf

. In this Demonstration, the amplitude

E

0

is set to 10 volts (V). You can vary the frequency

f

in Hz, the resistance

R

in ohms (

Ω

), the inductance

L

in millihenries (mH), and the capacitance

C

in microfarads (

μ

F).

X

L

=ωL

,

X

C

=

1

ωC

,



V

R

=R



I

,

V

L

=j

X

L



I

,



V

C

=-j

X

C



I

.

The phase of



V

R

is the same as that of



I

.



V

L

leads



I

by

0

90

. The phase of



V

C

lags that of



I

by

0

90

.

The voltage and current are out of phase by an angle

φ

, where

tanφ=

X

L

-

X

C

R

.

When

X

L

>

X

C

,

the effect of inductance is dominant; then

φ>0

, and the RLC circuit's total voltage



V

=



V

R

+



V

L

+



V

C

=(R+j

X

L

-j

X

C

)



I

leads the current



I

. When

X

L

<

X

C

,

the capacitance contribution is dominant,

φ<0

, and the current leads the voltage. When the circuit has a pure resistance or when the resonance condition

X

L

=

X

C

or

ω=

1

LC

is satisfied, then

φ=0

, meaning that the voltage and current are in phase.