báo cáo thí nghiệm môn mạch tuyến tính i

Kirchhoff''''s Current Law KCLKirchhoff''''s Current Law KCL states that the algebraic sum of currents leaving any node or the algebraic sum of currents entering any node is zero.. KCL can be

ĐẠI HỌC BÁCH KHOA HÀ NỘI

TRƯỜNG ĐIỆN- ĐIỆN TỬ

Lab #1: Kirchhoff's Current and Voltage laws

I OBJECTIVES

- To learn and apply Kirchhoff's Current Law.- To learn and apply Kirchhoff's Voltage Law.- To obtain further practice in electrical measurements.

- Compare experimental results with those using hand calculations.

II BACKGROUND AND THEORY1 Kirchhoff's Current Law (KCL)

Kirchhoff's Current Law (KCL) states that the algebraic sum of currents leaving any node or the algebraic sum of currents entering any node is zero KCL can be stated as the sum of the currents entering a node must equal the sum of the currents leaving a node.

Kirchhoff's Current Law can also be expressed as follows:

- The algebraic sum of the currents entering a junction (node) equals zero.- The algebraic sum of the currents leaving a junction (node) equals zero.- The algebraic sum of currents entering a node equal to the algebraic sum ofcurrents leaving a node.

2 Kirchhoff's Voltage Law (KVL)

Kirchhoff's Voltage Law (KVL) states that the algebraic sum of voltages around a closed path is zero As you make a summation of voltages, it is suggested that you proceed around the closed path in a clockwise direction If you encounter a positive (+) sign as you first enter the circuit element, add that value On the contrary, if you first encounter a negative sign as you enter the circuit element, then subtract the value of that voltage.

We can state Kirchhoff's Voltage Law in three ways which are all equivalent:

- The algebraic sum of the voltage drops around any closed path of an electric circuit equals zero.

- The algebraic sum of the voltage rises around any closed path of an electric circuit equal to zero.

- The algebraic sum of the voltage rises equals the algebraic sum of the voltage drops around any closed path of an electric circuit.

III EQUIPMENT AND PARTS LIST

- Digital Multimeter(DMM)- DC power supply- Circuit breadboard- Resistors

IV PROCEDURE

1 Without substituting in numbers for R1, R2, R3, and R4 apply Kirchhoff's Current Law at nodes 1 and 2 so as to obtain two equations in terms of the two unknown node voltages V1 and V2 Simplify these equations.

2 Let R1=100Ω, R2 = 200Ω, R3= 50Ω, and R4= 150Ω, voltage source Vs= 24V in the equations of Step 1 Solve these equations by hand for V1 and V2.

From V1 and V2 find Va, Vb, Vc, Ia, Ib, and Ic.

3 Measure the resistors Use these values to find V1, V2, Va, Vb, Vc, Ia, Ib, Ic as in Step 2.4 Construct the circuit shown in the Figure below:

5 Use the multimeter to measure the indicated three currents and five voltages6 Compare the results of Step 2 with those obtained from the theoretical calculationsof Steps 2 and 3 above.

7 Using the measured values of the three currents, check KCL at node 28 Use your measured values of the source voltage, Va Vb, and Vc to checkKirchhoff's Voltage Law for the outer loop of the circuit.

V CALCULATIONS AND COMPARISONS1 Calculations and comparisons

a Calculate and compare the current values

- Testing KCL

Ia+Ib−Ic=0- Applying Ohm’s Law, we have

Va−Vc=Vs=24(V )We obtain: I=71.111(mA ) ; I=−53.333(mA ; I) =17.777(mA )

24

- Applying Ohm’s Law, we obtain: V = 7.211 (V) ; V = 14.144 (V) ; V = 2.648 (V)abc

- To learn and apply Nodal analysis.

- To obtain further practice in electrical measurements.

- Compare experimental results with those using hand calculations.

II BACKGROUND AND THEORY

Analysis of electrical networks involves the determination of node voltages and loopor branch currents Nodal analysis refers to the technique of writing equations wherethe unknown quantities are the node voltages of the circuit Kirchhoff’s current law isused to define the equations at each node in the circuit, using currents obtained byOhm’s law.

III EQUIPMENT AND PARTS LIST

- Digital Multimeter (DMM)- DC power supply- Circuit breadboard- Resistors

IV PROCEDURE

Experiment 1:

1 Consider the circuit shown in Figure 1

2 Measure (with the multimeter) and record the resistance of each resistor you use Use the

multimeter to adjust the power supply.

3 Measure and record all node voltages and branch currents.4 Construct the circuit of Figure 1.

5 Measure (with the multimeter) and record the resistance of each resistor you use.

6 Measure and record all node voltages and branch currents.

- One useful property of the bridge circuit is the so-called balance condition that occurs when the relationship among the bridge resistors ("legs") is such that (R1/R2)-(R3/R4) Note that inthe balance state the node voltages V2 and V3 are equal, meaning that the current RS must bezero Also note that the balance condition depends only upon the resistor ratios, not the applied voltage V.

- The bridge circuit can be used to determine the value of an unknown resistor if several known resistors are available For example, if the R3 leg was an unknown resistor we could use various combinations of known resistors for the R1, R2, and R4 legs until the balance condition was achieved We would then know that R3- R4*(R1/R2).

- Using the resistance and voltages given in the Figure, determine all node voltages and branch currents in the circuit Using the voltage and current values, calculate the power dissipated by each resistor.

- Determine a new value for R4, leaving the other resistors unchanged, that will balance the bridge circuit Assume an adjustable resistor is used so that any resistance value can be obtained.

- Construct the circuit shown in the Figure.

- Measure and record the resistance of each resistor you use.

- Use the multimeter to adjust the power supply Measure and record all nodevoltages and branch currents.

Using node voltage method (NVM) to calculate the voltage through each component:

VB−VA=VS=24 V

We obtain:

V1=24 V; V2=12 V; Vc=12 VI1=0.12 A; I2=0.12 A; I3=0.12 A

2 Measured values

R1 200 Ω V1 24.271 V I1 0.114 A

R2 100 Ω V2 11.922 V I2 0.121 A

=> The measured data is very close to the calculated data The reason for the discrepancy is

the wire having little resistor and the inaccuracy of the measuring process.Experiment 2:

Im1=0.16 A; Im 2=−0.16 A ; Im 3=0.16 A

We can find the corresponding currents through each resistor:I1=0.32 A; I2=0.32 A; I3=0.16 A; I4=0.16 A; I5=0 AAnd the corresponding voltage:

V=16 V; V=8V ; V=16 V; V =8 V ; V=0V

- To build a Thévenin equivalent of the original circuit and check to see if it really is equivalent

II BACKGROUND & THEORY1 Thévenin's theorem

Thévenin's theorem states that any two-terminal circuit with linear elements can be represented with an equivalent circuit containing a single voltage source in series with a single resistor.

The equivalent circuit consists of an independent voltage source Voc and a series resistor R The resistor Rt is determined by removing all independent sources (short voltage sources and open current sources) Voc is found by measuring the open circuit voltage across the output and isc is found by measuring the current between the shorted output connections.Voc and Rt, are computed as follows: Vt = Voc

Thévenin's and Norton circuits are equivalent and if one is known, the other is also easily determined Actually, measuring the short circuit current in a real circuit is often not recommended (the circuit may not be designed to handle the high current) and may damage the circuit Circuits with no independent sources require a different technique; a source must be connected to the output and the current or voltage measured.

2 Maximum Power Transfer

The load resistance that absorbs the maximum power from a two terminal circuit is equal to the Thévenin resistance.

The maximum voltage at the output of a linear source is:Vmax=Voc=Vt=Rn

4 RT

4 =Voc

III EQUIPMENT AND PARTS LIST

- Digital Multimeter (DMM)- DC power supply- Circuit breadboard- Resistor

IV PROCEDURE

Experiment 1:

1 Assemble the circuit in Figure 4 without R

Use combinations of resistors from your kit to make R while assembling 5

the circuit Measure

the voltage across the point’s xy It is the open circuit voltage.

Short circuits the points x and y Now, measure the current flowing in the short-circuited branch ‘xy’ This gives you the short circuit current.Determine the Thevenin’s equivalent circuit from your measurements andcompare with the

theoretical data obtained earlier.

2 Remove the short circuit and place the resistor R of value equal to the L

Thevenin’s resistance found by measurements.

Use resistive decade box for obtaining this R Find the voltage across it L

and calculate the power dissipated in it Compare this power to the maximum power dissipated.

3 Now, construct the Thevenin equivalent network using the closest valueof resistor in your

parts kit for R Using a resistive decade box for R , connect it to the thL

Thevenin equivalentnetwork.

4 Vary R in increments of 500 from 0 to 10k For each value of RL L

record the load voltage.

Also record the load voltage at the theoretical value of R that maximizes L

the load power For

each value of load voltage calculate the load power.

As R is varied, make sure to do the measurement for R equal to R L L th

5 Modify the graphs from Theoretical Calculations, step 3 above by superimposing your

experimental results onto the theoretical graphs.

Use the plot symbol "o" to denote the experimental voltage or power

Don't connect the experimental data points by connecting lines End up with two graphs:

(a) Theoretical and experimental power versus RL

(b) Theoretical and experimental load voltage versus R over the range 0 L

< R < 20 k L

Experiment 2:

1 Construct the circuit below (using a 100 resistor with a large enough power rating to absorb isc) Measure voc and is at the output Remember to measure the resistance of each resistor used

Now imagine that you are unable to measure because a 100 resistor with a large enough power rating is unavailable (or pretend the power supply might be damaged).

Take an additional measurement that will allow you to construct the Thevenin equivalent.2 Place the following "loads" across the output and record voltage

across and current through.(Use 50Ω, 100Ω, 150Ω and 200Ω)

V EXPERIMENT RESULTS

Vopen = 18.82 (V)Ishort = 102.65 (mA)

Equivalent Thevenin’s resistor: 183.33 (Ω)When loads are placed:

voltage will be the voltage across the load when no current is flowing, and the Thevenin resistance will be the ratio of the voltage to the current when the load is connected.- With this additional measurement data, we can construct the Thevenin equivalent circuit, which consists of a voltage source (Vth) in series with a resistor (Rth) The Thevenin voltage will be the open-circuit voltage measured earlier, and the Thevenin resistance will be calculated based on the load measurements.

Lab #6: Superposition

I OBJECTIVES

- Study the principle of superposition.

- Analyze the circuits used in this experiment numerically in the preliminary lab exercise

- Examine experimentally same circuits in the lab and the experimental values compared to

the theoretical.

II BACKGROUND & THEORY

- A circuit component is called linear if the current through the componentis linearly proportional to the voltage across the component or in another word the I-V relationship is linear.

- The superposition theorem is applied to a simple linear circuit containingmore than one source The experiment will show superposition can be applied to voltage and current but not to power Also, a battery is used as a power source to demonstrate a non-ideal voltage source.

- The superposition principle states that the total response is the sum of the responses to each of the independent sources acting individually Superposition principle does not apply to any circuit that has element (s) described by nonlinear equation (s).

- Time-varying currents having different frequencies will flow through linear circuits

independently hence no new frequencies will be produced in the circuits.- Generally, resistors, capacitors, inductors and transformers are linear components if they operate at suitably low currents and voltages Circuit components that behave linearly at low currents and voltages may behave nonlinearly if subjected to extreme currents and voltages.III EQUIPMENT AND PARTS LIST

- Digital Multimeter (DMM)- DC power supply- Circuit bread board- Resistors- Signal Generator- OscilloscopeIV PROCEDURE

Experiment 1:

1 Consider the circuit in Figure 1 Using node voltage analysis, calculate the voltage V’ across 3.3 k resistor and the current flowing through it.2 Now use the principle of superposition Find the voltage across 3.3k due to individual

voltage sources and calculate the voltage V’ Find the current flowing through 3.3k due to

individual sources and the net current through it.

3 Assemble the circuit and measure the voltage V’ and the current through 3.3k

4 Remove one of the voltage sources in the circuit Replace it with a shortcircuit Measure the voltage across 3.3k and current through it due to this source.

5 Now remove the short circuit and reconnect this source in its place in the circuit.

6 Repeat Step 4 with the other source.

7 Add the voltages measured from Step 4 and 6 keeping in mind the polarities of voltages

measured and direction of current measure.

8 Compare with the theoretical calculations and comment on the results.Experiment 2:

Consider the circuits shown in Figure 2.

1 Solve the circuit shown in Figure 2a for the voltages V , V , and V and 123

for the currents I , I and I Use either the node-voltage or mesh-current 12, 3

method (loop analysis) Do not use theprinciple of superposition.

2 Suppress (zero) V in the circuit of Figure 2a by replacing it with a shortS2

circuit Solve the

resulting circuit, shown in Figure 2b, for the voltages V ’, V ’, and V ’ and 123

for the currents I1’,I2’, and I ’.3

3 Suppress (zero) V in the circuit of Figure 2a by replacing it with a shortS1

Solve the resulting circuit, shown in Figure 2c, for the voltages V ”, V ”, 12

and V ” and for the3

4 Combine the results of step 2 and step 3 using the principal of superposition and compare with the solution of step 1.V EXPERIMENT RESULTS

Experiment 1:Experiment 2:1 Measured valuesa Both 24V and 12V source:

After some calculations, we obtain:

a Both 24V and 12V source:

a Both 24V and 12V source:

Measured values(V) Calculated values(V)Differences(%)

II BACKGROUND AND THEORY

The characteristic equation is derived as

A convenient way to examine the characteristic equation is to compare a given secondorder characteristic equation with a standard form expressed as

ζ is called the damping factor: ω is called the undamped natural resonant frequency If ζ < 1 the system response is underdamped

ζ = 1 the system response is critically

ζ > 1 the system response is overdamped

III EQUIPMENT AND PARTS LIST

- Digital multimeter (DMM)- DC power supply- Circuit breadboard- Resistors- Pulse generator- Oscilloscope- Capacitor- Inductor

IV EXPERIMENT RESULTS1 Circuit for experiment

- Actual experiment data:

50Ω

V CONCLUSIONS

This experiment shows the oscillation in the second order circuits as we have mentioned in the background and theory As we change the resistor, the nature of the circuit is changing between overdamped, critically damped and underdamped responses The type of response ofthe circuit will result in different waveforms of capacitor voltage.