Experiment no.
2
R L C series circuits
Object:
To study the characteristics of ac circuits.
Apparatus:
1- Dual beam oscilloscope.
2- Function generator.
3- Resistance box.
4- Capacitance box.
5- Inductance box.
6- Coaxial cable.
7- Connecting wires.
Theory:
A- Impedance : the impedance of a two terminal network may be
expressed as:
Z=v/I (ohms)
Where z = complex impedance
V= complex voltage
I complex current
The complex impedance is also expressed as:
Z=r+jx in Cartesian form
Z=zeiФ in polar form.
B- THE SERIES RC CIRCUIT:
Fig 2.1 illustrate a series rc circuit connected to an ac voltage
source.
The applied voltage can be expressed as:
V=Vm sinɷt
=Ir +1/C∫ 𝑖𝑑𝑡
The solusion of this differential equation results in.
I=IM sin(ɷt+Ф)
This equation shows that current I leads voltage v by Ф
The impressed voltage v can expressed form as:
V=VR-jVC
Which can be expressed as(see phasor diagram):
|v|= 𝑉 +𝑉
= (𝐼𝑅) + (𝐼 𝑋 )
=|I| (𝑅) + ( 𝑋 )
Where XC=1/(ɷC)
And the phase angle Ф is given by:
Ф=tan-1(-VC/VR)=tan-1(-XC/R)
a) Circuit and wave form
Fig 2.1 series RC
b) phasor diagram
FIG. 2.1 series RC
C- Series RL circuit:
Fig.2.2 illustrate a series RL circuit to which a sinusoidal voltage
v=VM sin ɷt is
impressed.
Applying K.V.L, one gets:
V=IR-Ldi/dt
Solving eqn for I, one gets
I=IM Sin(ɷt-Ф)
Thus the current I lags the voltage v by Ф.
From phasor diagram, one express:
|v|= 𝑉 +𝑉
= (𝐼𝑅) + (𝐼 𝑋 )
=|I| (𝑅) + ( 𝑋 )
Where XL= ɷL
Ф=tan-1(VL/VR)=tan-1(XL/R)
a) circuit and waveforms
b) Phasor diagram
Fig 2.2 a series RL circuit
D- THE SERIES RLC CIRCUIT:
A series RLC circuit is illustrated in fig.2.3. as obvious from the
figure the effect of XL and XC are opposite. VL leads the current by
90° while VC lags the current by 90° thusthe voltage across capacitor
and inductor are out of phase by 180°. If the magnitude of these
voltage are equal, they cancel each other and the total reactance in the
circuit
XL-XC=0
The impedance of the circuit is
Z=R+jXL-jXC
And the phase angle Ф is given by
Ф=tan-1((VL-VC)/VR)
=tan-1((XL-XC)/ R)
If XL=XC the circuit is resistive
XL>XC the circuit is inductive
XL<XC the circuit is capacitive
a) Circuit b) phasor diagram
c)impedance triangle
fig.2.3.A series RLC circuit
Procedure
Part A: series RC circuit
i- Connect the circuit shown in fig 2.1
ii- Set the input voltage at 3VP.P, 300Hz
iii- Measure the phase shift between the current I and the applied
voltage v by using oscilloscope.
iv- Calculate the phase angle theoretically.
Part B:series RL Circuit
i- Connect the circuit shown in fig.2.2
ii- Set input voltage at 4VP.P, 250Hz
iii- Repeat step (iii)&(iv) in part A.
Part C: series RLC circuit
i- Connect the circuit shown in fig.2.3
ii- Set the input voltage at 4 VP.P 1 kHz
iii- Measure the phase shift between input voltage and total current.
iv- Calculate the phase angle theoretically
Discussion
1- Explain why the phasor and the impedance have the same angle.
2- At what condition the following results obtained?
a- Phase angle equal zero
b- The applied voltage lead the current by 90°.
c- The average power equal to zero.
3- In general, how would the phasor diagram of Figure 2.1 change if
the frequency was raised?
4- In general, how would the phasor diagram of Figure 2.2 change if
the frequency was lowered?
