Electrical Circuit Lab1(7)

8/2/2019 Electrical Circuit Lab1(7)

1/8

Objective

- In this experiment the natural and step responses of RC and RL circuits are

examined.

- The use of computer controlled equipment is also introduced here.

Theory

- Introduction and test circuits:

Inductors and capacitors have the ability to store energy. It is important to

determine the voltages and currents that arise in circuits composed by resistors.

And either inductors or capacitors. The description of the voltages and currents

in this type of circuits is done in terms of differential equations of first order.

- Natural Response:

The currents and voltages that arise when the energy stored in an inductor or

capacitor is suddenly released to the resistors in the circuit are called the natural

response of the circuit. The behavior of these currents and voltages depends only

on the nature of the circuit, and not on external sources.

1. Natural response of an RL circuit

In an RL circuit, the natural response is described in terms of the voltages and

current at the terminals of the resistor when the external source of power stops

delivering energy to the circuit. The expressions for the current and voltageacross the resistor are:

So,

, .

Where I0 is the initial current through the inductor before the power source

goes off and the inductor starts releasing energy to the circuit.

The symbol represents the time constant of the circuit.

The natural response of an RL circuit is calculated by:

- Finding the initial current I0 through the inductor- Finding the time constant of the circuit

- Generate i(t).

2. Natural response of an RC circuit

The natural response of an RC circuit is analogous to that of an RL circuit. The

expressions for the current and voltage across the resistor are;

So, , .

Where V0 is the initial voltage across the (fully charged) capacitor before thepower source is switched off, the capacitor starts releasing energy to the circuit.

The natural response of an RC circuit is calculated by:


2/8

- Finding the initial voltage V0 across the capacitor

- Finding the time constant of the circuit

- Generating V(t)

- Parallel RLC circuit:The current in the circuit consisting of a resistor, a capacitor and an inductor

connected in parallel is given by:

Where is given by: 12RC

The resonant radian frequency in (rad/sec), given by:o

1

LC

Equipment and instruments

- Digital Multimeter (DMM).

- The Function Generator (FG).

- The Cathode Ray Oscilloscope (CRO).

- Various components.

Damping Natural response equations

Overdamped

2 > 2 .1 2( )

1 2S t S t

i t A e A e 2 o1,2

s

Underdamped

2 < 2

t( ) [ B cos t + B sin t ]1 2d d

i t e

2 20

w wd

Critically

damped

2=2

t( ) [ A t +A ]1 2

i t e

A1,A2 are constantsdetermined b the initial

condition


3/8

Procedure

1. Step response of RL Circuits:

a. DC RL Circuit.

1- Assemble the circuit in the previous figure with the component values. Byusing Ohmmeter measure the internal resistance of the Inductor.

2- Take measurement to find the experimental values for the components, the

current flowing through the inductor and the voltage across each element.

b. Transient RL circuit:

1- Assemble the circuit in the previous figure with the component values.

Measure the internal resistance of the inductor.

2- Set the FG to supply a square wave with 6VPK-PK amplitude and 10 KHz

frequency. Add 3V DC offset. Check the amplitude of the signal after the

connection to the circuit.3- Connect CH1 and CH2 as illustrated on the graph, notice that there is a small

resistance R2 connected in series with the inductor. This is because the


4/8

Oscilloscope can measure voltage only, therefore its necessary to measure scaled

version of ic(t).

Divide the voltage by R to obtain the current in the circuit,

So CH2 represented iL(t).

4- Observe the waveform of voltage and current, and plot them on a graphpaper.

5- Calculate the time constant for this circuit and compare with the voltage

and current at t= , and t=3.

2. RC circuits:

a. DC RC circuit


2- Take measurement to find the values of the resistors, the capacitor and the

voltages across them.


5/8

b. Transient RC circuit


2- Set FG to supply a square wave with 4 VPK-PK amplitude

And 10 KHz frequency. Add 2V DC offset. Check the amplitude of the signal afterthe connection to the circuit.

3- Connect CH1 and CH2 as illustrated on the graph, notice that there is a

small resistance R2 connected in series with the capacitor. This is because the

Oscilloscope can measure voltage only, therefore, its necessary to measure

scaled version of ic(t). Divide the voltage by R to obtain the current in the circuit,

So CH2 represent ic(t).

4- Observe the waveform of voltage and current, and Plot them on a graph

paper.

5- Calculate the time constant for this circuit and compare with the voltage

and current at t=, and t= 3.

2V

R1

1kohm

R2

10ohm

C

10nF

CH1

CH2

Common

Ground

DC offest

4Vp-p

10 KHz


6/8

Data & Calculation

1. Step response of RL Circuits:

a. DC RL Circuit.

Parameter Unit Theor. Exp. %ErrorR1 K 2.2 2.18 1.66%

RL(internal resistance ofinductor)

--- 120

L mH 47 --- 0%

Vs

V

10 9.98 0.1 %

VR1 10 9.68 5.2%

VL 0 0.335

IL

mA 8.33 4.25 4.92 %

b. Transient RL circuit:

-The current of the inductor

- The Voltage of the inductor


7/8

iL () = I0(1-e-t/)

= 63.21 mA , iL(3)=I0(1-e

-3) = 95.02 mA

VL()=Vs e-t/

= 3.67 V ,VL(3) = Vse-3

= 0.5 V

2. RC circuits:a. DC RC circuit

Parameter Unit Theor. Exp. %Error

R1

K

2.2 2.16 1.66%

R2 1.2 1.18 1 %

C F 100 --- 0 %

Vs

V

10 9.98 0.1 %

VR1 5.455 6.44 0.2 %

VR2 4.545 3.48 0.3 %

VC 4.545 3.48 0.3 %

b. Transient RC circuit

-The current of the capacitor

- The Voltage of the Capacitor.


R1 K 1.0 0.99 1 %

R2 10 9.65 3.5 %

Rint 0 98.33L mH 10 10 0 %

sec 10 *10-6

9*10-6

10 %


8/8

Vc() = Vs(1-e-t/)

=2.78 V , Vc(3) =Vs (1-e

-3) = 4.8 V

Ic ()=I0 e-t/

= 14.72 mA , Ic(3) = I0e-3

= 2 mA

Conclusion

1) We conclude that each circuit has a natural response & its different due to

the circuit elements.

2) There is no pure inductive or capacitive load, without any internal resistanceand we should consider it in our calculations.

3) The inductor and the capacitor are none absorbing power elements on the

contrary they are elements that storage power, the inductor storages the power

on the shape of electromagnetic waves and the capacitor on the shape of

charges.

4) The values of and02

determine the form of the natural (or step) response

of parallel RLC circuits.

5) Depending on the damping, the solution to the differential equation

describing the response of the circuit can be found by applying the appropriate

set of equation.


R1 K 1.0 0.99 1 %

R2 10 10 0 %

C nF 10 --- 0 %

Sec 10.01 *10-6

10*10-6

0 %

Documents

Electrical Circuit Lab1(7)