báo cáo tn mạch i báo cáo thí nghiệm tn mạch tuyến tính 1 ee3716

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

ĐẠI HỌC BÁCH KHOA HÀ NỘI TRƯỜNG ĐIỆN ĐIỆN - TỬ

Hà Nội, ngày 29, tháng 5, 2024.

Lab #1: Kirchhoff's Current and VoltageLaws

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 THEORY 1 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 follows: as

- 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 of

currents 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 val ue On the contrary, if you first encounter a negative sign as you enter the circuit elem ent, 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

2 Let R1=100Ω, R2 = 500Ω, R3= 50Ω, and R4= 150Ω 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 lc.

3 Measure the resistors Use these values to find V1, V2, Va, Vb, Vc, Ia, Ib, and 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 voltages

6 Compare the results of Step 2 with those obtained from the theoretical calculations of Steps 2 and 3 above

7 Using the measured values of the three currents, check KCL at node 2

8 Use your measured values of the source voltage, Va Vb, and Vc to check Kirchhoff's Voltage Law for the outer loop of the circuit

V CALCULATIONS AND COMPARISONS 1 Actual Values

Voltage (V) Current (I) Va = 7.06 (V) Ia = 69.68 (mA) Vb = 14.35 (V) Ib = 51.34 (mA) Vc = 2.53 (V) Ic = 17.42 (mA)

VI CONCLUSIONS

The measured value are reasonably accurate and are roughly the same as the nominal values

Lab #2: Nodal Analysis I OBJECTIVES

- 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 loop or branch currents Nodal analysis refers to the technique of writing equations where the unknown quantities are the node voltages of the circuit Kirchhoff's current law is used to define the equations at each node in the circuit, using currents obtained by Ohm's law

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

IV PROCEDURE

- Consider the bridge circuit shown in Figure below

- 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 in the balance state the node voltages V2 and V3 are equal, meaning that the current RS must be zero 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 node voltages and branch currents

V EXPERIMENTAL RESULTS 4 Actual Values

6 Measured values

Voltage (V) Current (I) V1 = 15.84 (V) I1 = 322.61 (mA)

V2 = 7.62 (V) I2 = 331.74 (mA) V3 = 15.72 (V) I3 = 155.60 (mA) V4 = 8.06 (V) I4 = 172.68 (mA)

Lab #4: Thévenin and Norton EquivalentCircuits

I OBJECTIVES

- To understand the Thévenin's and Norton equivalent of a circuit - To check the experimental values versus calculated values - To find the conditions for maximum power delivered to a load

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

II BACKGROUND & THEORY 1 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) V is found by measuring the open circuit oc

voltage across the output and i is found by measuring the current between the sc

shorted output connections

Voc and R , t are computed as follows: V = Vtoc

Thévenin's and Norton circuits are equivalent and if one known, the other is also iseasily 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:

The maximum current at the output of a linear source is:

The maximum power delivered by a linear source to matched load resistance R L= R = R Tnis:

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

IV PROCEDURE

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

EX2 part1 :

Theoretical calculation Measured value Rth = 97 ( ) Ω

Vth = 11.52 (V) Vth = 11 (V) Ith = 0.118 (A) Ith = 0.12 (A)

Actual value: R 100 -> 50 ΩR 220 -> 100 ΩR 330 -> 150 Ω

b When V = 12 (V)

Theoretical calculation Measured valueRth = 183.33 (Ω)

Vth = 8 (V) Vth = 8.06 (V) Ith = 0.0437 (A) Ith = 0.044 (A)

EX2 part2 :

Actual value: V = 12V Rth = 250

Theoretical calculation Measured valueRth = 250 (Ω)

Vth = 10.9 (V) Vth = 11 (V) Ith = 0.0436 (A) Ith = 0.044 (A)

VI CONCLUSIONS

By measuring the voltage and current for each load, we can determine the equivalent Thevenin voltage (Vth) and the Thevenin resistance (Rth) of the circuit The Thevenin 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 component is linearly proportional to the voltage across the comp onent or in another word t he I-V relationship is linear

The superposition theorem is applied to a simple linear circuit containing more 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 theyoperate 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

• Digital Multimeter (DMM)

• DC power supply

50

volts200 24

volts • Circuit bread board

Figure 1 An example of circuit for 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 short circuit 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 Figure 2a in for the voltages V , V , and V and for the 1 2 3 currents I1 2, I , and I Use either the node-voltage or mesh-3 current method (loop analysis) Do not use the principle of superposition

I1 (theory) I2 (theory) I3 (theory) I1(practical) I2 (practical) I3 (practical) V = 24 (V) 0.21 (A) 0.135 (A) 0.075 (A) 0.3 (A) 0.13 (A) 0.17 (A) V = 12 (V) 0.034 (A) -0.017 (A) 0.051 (A) 0.035 (A) -0.016 (A) 0.054 (A)

V = 24,12

(V) 0.244 (A) 0.118 (A) 0.126 (A) 0.335 (A) 0.114 (A) 0.224 (A) I (theory) I (practical)

V = 24 (V) 0.087 (A) 0.09 (A) V = 12 (V) 0.022 (A) 0.03 (A) V1 = 24 (V), V2 = 12

24 volts24 volts

12 volts

12 volts

2 Suppress (zero) VS2 in the circuit of Figure 2a by replacing it with a short circuit Solve the resulting circuit, shown in Figure 2b, for the voltages V1’, V2’, and V3’ and for the currents I1’, I2’, and I3’

3 Suppress (zero) VS1 in the circuit of Figure 2a by replacing it with a short circuit Solve the resulting circuit, shown in Figure 2c, for the voltages V1”, V2”, and V3” and for the currents I1”, I2”, and I3”

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

Lab #8: Second Order Circuit I OBJECTIVES

Tostudy the step response of second order circuits.To understand the difference between overdamped, critically damped, and underdamped responses

To determine theoretically and experimentally the damped natural frequency is the underdamped case

II BACKGROUND AND THEORY

The characteristic equation derived is as

A convenient way examine to the characteristic equation is to compare a given second- order 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 critically isζ > 1 the system response overdamped is

With (1)and(2), solving the equation we have:

- Digital multimeter (DMM) - DC power supply - Circuit breadboard

- Resistors - Pulse generator - Oscilloscope - Capacitor - Inductor

IV EXPERIMENT RESULTS 1 Circuit for experiment

- Actual experiment data:

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 of the circuit will result in different waveforms of capacitor voltage