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10 Series-Parallel Combination Circuit OBJECTIVES: After performing this experiment, you will be able to: 1

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10 Series-Parallel Combination Circuit

OBJECTIVES:

After performing this experiment, you will be able to:

1. Use the concept of equivalent circuits to simplify series-parallel circuit analysis.

2. Compute the currents and voltages in a series-parallel combination circuit and verify your computation with circuit measurements.

READING:

Floyd, Principles of Electric Circuits, Sections 7-1 through 7-7, and TECH TIP

MATERIALS NEEDED:

Resistors:

One 2.2 kW, one 4.7 kW, one 5.6 kW, one 10 kW

SUMMARY OF THEORY:

Most electronic circuits are not just series or just parallel circuits. Instead, they may contain combinations of components. Many circuits can be analyzed by applying the ideas developed for series and parallel circuits to them. Remember that in a series circuit the same current flows through all components and that the total resistance of series resistors is the sum of the individual resistors. By contrast, in parallel circuits, the applied voltage is the same across all branches and the total resistance is given by the reciprocal formula.

In this experiment, the circuit elements are connected in composite circuits containing both series and parallel combinations. The key to solving these circuits is to form equivalent circuits from the series or parallel elements. You need to recognize when circuit elements are connected in series or parallel in order to form the equivalent circuit. For example, in Figure 10—1(a) we see that the identical current is in both R₂ and R₃. We conclude that these resistors are in series and could be replaced by an equivalent resistor equal to their sum. Figure 10—1(b) illustrates this idea. The circuit has been simplified to an equivalent parallel circuit. After finding the currents in the equivalent circuit, the results can be applied to the original circuit to complete the solution.

V_s   R₁   R₂     R₃    (a)                             V_s    R₁                R_2.3     (b)

Figure 10-1

The answer to two questions will help you identify a series or parallel connection: (1) Will the identical current go through both components? If the answer is yes, the components are in series. (2) Are both ends of one component connected directly to both ends of another component? If yes, the components are in parallel. The components that are in series or parallel may be replaced with an equivalent component. This process continues until the circuit is reduced to a simple series or parallel circuit. After solving the equivalent circuit, the process is reversed in order to apply the solution to the original circuit. This idea will be studied in this experiment.

PROCEDURE:

1. Measure and record the actual values of the four resistors listed in Table 10-1,

2. Connect the circuit shown in Figure 10-2. Notice that the identical current is through R₁ and R₄, so we know that they are in series. R₂ has both ends connected directly to R₃, so these resistors are in parallel.

V_s = +12 V       R₁=2.2 kW        R₂=4.7 kW      R₃=5.6 kW        R₄=10.0 kW

Figure 10-2 | | >

3. You can begin solving for the currents and voltages in the circuit by replacing resistors that are either in series or in parallel with an equivalent resistor. In this case, begin by replacing R₂ and R₁  with one equivalent resistor. Label the equivalent resistor R_2.3. Draw the equivalent series circuit in the space provided in the report. Show the value of all components, including R_2.3.

4. The equivalent circuit you drew in step 3 is a series circuit. Compute the total resistance of this equivalent circuit and enter it in the first two columns of Table 10-2. Then disconnect the power supply and measure the total resistance to confirm your calculation. Enter the measured total resistance, R_T, in column 3.

5. The voltage divider rule can be applied directly to the equivalent series circuit to find the voltages across R₁, R_2.3, and R₄. Find V₁, V_2.3, and V₄ using the voltage divider rule. Tabulate the results in

Table 10-2 in the Voltage Divider column.

6. Find the total current, I_T, in the circuit by substituting the total voltage and the total resistance into Ohm’s law. Enter the computed total current in Table 10—2 in the Ohm's law column.

7. In the equivalent series circuit, the total current is through R₁, R_2.3, and R₄. The voltage drop across each of these resistors can be found by applying Ohms law to each resistor. Compute V₁, V_2.3, and V₄ using this method. Enter the voltages in Table 10—2 in the Ohm’s law column.

8. Use V_2.3, and Ohm’s law to compute the current in R₂ and R₃ of the original circuit. Enter the computed current in Table 10-2. As a check, verify that the computed sum of I₂ and I ₃ is equal to the computed total current.

9. Measure the voltages V₁, V_2.3. V₄, and V_s. Enter the measured values in Table 10-2.

10). Change the circuit to the circuit shown in Figure 10-3. In the space provided in your report, draw an equivalent circuit by combining the resistors that are in series. Enter the values of the equivalent resistors on your schematic drawing and in Table 10-3.

V_s=+12 V      R₁=2.2 kW      R₂=4.7 kW    R₃=5.6 kW           R₄=10.0 kW

Figure 10-3

11. Compute the total resistance, R_T, of the equivalent circuit. Then apply Ohm’s law to find the total current I_T. Enter the computed resistance and current in Table 10-3.

12. Complete the calculations of the circuit by solving for the remaining currents and voltages listed in Table 10-3. Then measure the voltages across each resistor to confirm your computation.

FOR FURTHER INVESTIGATION:

Figure 10-4 illustrates another series-parallel circuit using the same resistors. Develop a procedure for solving the currents and voltages throughout the circuit. Summarize your procedure in a laboratory report. Confirm your method by computing and measuring the voltages in the circuit.

V_s=+12 V      R₁=2.2 kW      R₂=4.7 kW    R₃=5.6 kW           R₄=10.0 kW

Figure 10-4

Report for Experiment 10

ABSTRACT:

DATA:

Table 10-1

Component	Listed Value	Measured Value
R1	2.2 kW
R2	4.7 kW
R3	5.6 kW
R4	10.0 kW

 

Step 3 Equivalent Series Circuit

 

Table 10-2

	Computed		Measured
	Voltage Divider	Ohm’s Law
RT
IT
VI
V2.3
V4
I2
I3
VS	12.0 V	12.0 V

 

20 Capacitors

OBJECTIVES:

After performing this experiment, you will be able to:

1. Compare total capacitance, charge, and voltage drop for capacitors connected in series and in parallel.

2. Test capacitors with an ohmmeter and a voltmeter as a basic charging test.

READING:

Floyd, Principles of Electric Circuits, Sections 13-1 through 13-5 and 13-8

MATERIALS NEEDED:

Two LEDs

Resistors: Two 1.0 kW

Capacitors: One of each: 100 mF, 47 mF, 1.0 mF, 0.1 mF, 0.01 mF (35 WY or greater)

Application Problem: One additional 100 mF capacitor, one 100 kW resistor

SUMMARY OF THEORY:

A capacitor is formed whenever two conductors are separated by an insulating material. When a voltage exists between the conductors, there will be an electric charge between the conductors. The ability to store an electric charge is a fundamental property of capacitors and affects both de and ac circuits. Capacitors are made with large flat conductors called plates. The plates are separated with an insulating material called a dielectric. The ability to store charge increases with larger plate size and closer separation.

When a voltage is connected across a capacitor, charge will flow in the external circuit until the voltage across the capacitor is equal to the applied voltage. The charge that flows is proportional to the size of the capacitor and the applied voltage. This is a fundamental concept for capacitors and is given by the equation:

                               Q = CV

Where Q is the charge in coulombs, C is the capacitance in farads, and V is the applied voltage. An analogous situation is that of putting compressed air into a bottle. The quantity of air is directly proportional to the capacity of the bottle and the applied pressure.

Recall that current is defined as charge per time; that is,

                                I  = Q/t

Where I  is the current in amperes, Q is the charge in coulombs, and t is the time in seconds. This equation can be rearranged as

                                            Q = I t

If we connect two capacitors in series with a voltage source, the same charging current flows through both capacitors. Since this current flows for the same amount of time, it can be seen that the total charge, Q_T, must be the same as the charge on each capacitor; that is,

Q_T = Q₁ = Q₂

Charging capacitors in series causes the same charge to be across each capacitor: however, the total capacitance decreases. In a series circuit, the total capacitance is given by the formula

1/C_T = 1/C₁ + 1/C₂ + …… + 1/C_i

Now consider capacitors in parallel. In a parallel circuit, the total current is equal to the sum of the currents in each branch as stated by Kirchhoff’s current law. If this current flows for the same amount of time, the total charge leaving the voltage source will equal the sum of the charges which flow in each branch. Mathematically,

               Q_T = Q₁ + Q₂ + ….. + Q_i

Capacitors connected in parallel will raise the total capacitance because more charge can be stored at a given voltage. The equation for the total capacitance of parallel capacitors is:

                   C_T = C₁ + C₂ + ….. + C_i

There are two quick tests that can verify that a capacitor, larger than about 0.01 mF, can be charged. Although the two tests are not comprehensive, they are useful in troubleshooting a faulty) capacitor. The first test uses only an ohmmeter as a visual indication of charging to a small voltage (the internal voltage in the ohmmeter). An analog ohmmeter is best for this test. This test 1s done as follows:

1. Remove one end of the capacitor from the circuit and discharge it by placing a short across its terminals.

2. Set the ohmmeter on a high resistance scale and place the negative lead from an ohmmeter on the negative terminal of the capacitor. You must connect the ohmmeter with the proper polarity. Do not assume the common lead from the ohmmeter is the negative side!

3. Touch the other lead of the ohmmeter onto the remaining terminal of the capacitor. The meter should indicate very low resistance and then gradually increase resistance. If you put the meter in a higher range, the ohmmeter charges the capacitor slower and the capacitance “kick’’ will be emphasized. For small capacitors (under 0.01 mF), this change may not be seen. Large electrolytic capacitors require more time to charge, so use a lower range on your ohmmeter. Capacitors should never remain near zero resistance, as this indicates a short. An immediate high- resistance reading indicates an open for larger capacitors.

A capacitor that passes the ohmmeter test may fail when working voltage 1s applied. A voltmeter can be used to check a capacitor with voltage applied. The voltmeter is connected in series with the capacitor and a de voltage as indicated in Figure 20—1. When voltage is first applied, the capacitor charges through the voltmeter’s large series resistance. As it charges, voltage will appear across it, and the

+35 V                        V

Figure 20-1

Voltmeter indication will soon show a very small voltage. Large electrolytic capacitors may have leakage current that makes them appear bad, especially with a very high impedance voltmeter. In this case, use the test as a relative test, comparing the reading with a similar capacitor which you know is good.

The simple charging tests are satisfactory for determining if a gross failure has occurred. They do not indicate the value of the capacitor or if its value has changed. Value change is a common fault in capacitors, and there are other failures, such as high leakage current and dielectric absorption (the result of internal dipoles remaining in a polarized state even after the capacitor discharges). Some low cost DMMs include built-in capacitance meters. A more comprehensive test can be provided by an instrument such as a dynamic component analyzer, which measures the value as well as leakage current and dielectric absorption.

Capacitor Identification

There are many types of capacitors available with a wide variety of specifications for size, voltage rating, frequency range, temperature stability, leakage current, and so forth. For general-purpose applications, small capacitors are constructed with paper, ceramic, or other insulation material and are not polarized. Three common methods for showing the value of a small capacitor are shown in Figure 20-2. In Figure 20—2(a), a coded number is stamped on the capacitor that is read in pF. The first two digits represent the first 2 digits, the third number is a multiplier. For example, the number 473 is a 47000 pF capacitor. Capacitors under 100 pF will not include a multiplier digit. Figure 20-2(b) shows the actual value stamped on the capacitor in mF. In the example shown, .047 mF is the same as 47000 pF. In Figure 20-2(c), a ceramic color-coded capacitor is shown that is read in pF. Generally, when 5 colors are shown, the first is a temperature coefficient (in ppm/°C with special meanings to each color). The second, third, and fourth colors are read as digit 1, digit 2, and a multiplier. The last color is the tolerance. Thus a 47000 pF capacitor will have a color representing the temperature coefficient followed by yellow, violet, and orange bands representing the value. Unlike resistors, the tolerance band is generally green for 5% and white for 10%.

Manufacturer’s code      First digit   Second digit    Multiplier   ®  473 = 47000 pF

            XX473

1. Coded value

   .047     ¬       Value

   XXX     ¬   Other information may be included

2. Stamped value (in pF or mF)

Temperature coefficient    First digit   Second digit     Multiplier    Tolerance

4700 pF = yellow violet orange in center three positions

3. Color code

Figure 20-2

Larger electrolytic capacitors will generally have their value printed in uncoded form on the capacitor and a mark indicating either the positive or negative lead. They also have a maximum working voltage printed on them which must not be exceeded. Electrolytic capacitors are always polarized, and it is very important to place them into a circuit in the correct direction based on the polarity shown on the capacitor. They can overheat and explode if placed in the circuit backwards. 7

PROCEDURE:

1. Obtain five capacitors as listed in Table 20-1. Check each capacitor using the ohmmeter test described in the Summary of Theory. Record the results of the test on Table 20-1.

2. Test each capacitor using the voltmeter test. Because of slow charging, a large electrolytic capacitor may appear to fail this test. Check the voltage rating on the capacitor to be sure it is not exceeded. The working voltage is the maximum voltage that can safely be applied to the capacitor. Record your results in Table 20-1.

3. Connect the circuit shown in Figure 20-3. The switches can be made from jumper wires. Leave both switches open. The light-emitting diodes (LEDs) and the capacitor are both polarized components—they must be connected in the correct direction in order to work properly.

R₁ 1.0         S₂   R₂ 1.0 kW   +12 V  S₁    C₁  100 mF   

Figure 20-3

4. Close S₁ and observe the LEDs. Describe your observation in Table 20-2.

5. Open S₁ and then close S₂. Describe your observations in Table 20-2.

6. Now connect C₂ in series with C₁. Open both switches. Make certain the capacitors are fully discharged by shorting them with a piece of wire; then close S₁. Measure the voltage across each capacitor. Do this quickly to prevent the meter from causing the capacitors to discharge. Record the voltages in Table 20—2.

7. Using the measured voltages, compute the charge on each capacitor. Then open S₁ and close S₂. Record the computed charge and your observations in Table 20-2.

8. Change the capacitors from series to parallel. Open both switches. Ensure the capacitors ate fulls discharged. Then close S₁. Measure the voltage (quickly) across the parallel capacitors and enter the measured voltage in Table 20-2.

9. Using the measured voltage across the parallel capacitors, compute the charge on each one. Then open S₁ and close S₂. Record the computed charge and your observations in Table 20-2.

10. Replace the +12 V dc source with a signal generator. Set the signal generator to a square wave and set the amplitude to 12 V_pp. Set the frequency to 10 Hz. Close both switches. Notice the difference in the LED pulses. This demonstrates one of the principal applications of large capacitors—that of filtering. Record your observations.

FOR FURTHER INVESTIGATION:

Use the oscilloscope to measure the waveforms across the capacitors and the LEDs in step 10. Try speeding up the signal generator and observe the waveforms. Use the two-channel-difference measurement to see the waveform across the ungrounded LED (connect one channel to each side of the LED and select CH1-CH2). Draw and label the waveforms on the plots provided in the report.

APPLICATION PROBLEM:

A voltage multiplier is a circuit that uses diodes and capacitors to increase the peak value of a sine wave. Voltage multipliers can produce high voltages without requiring a high-voltage transformer. The circuit illustrated in Figure 20-4 is a full-wave voltage doubler. The circuit is drawn as a bridge with diodes in two arms and capacitors in two arms. The diodes allow current to flow in only one direction, charging the capacitors to near the peak voltage of the sine wave. Generally, voltage doublers are used with 60 Hz power line frequencies and with ordinary diodes, but in order to clarify the operation of this circuit, you can use the LEDs that were used in this experiment. (Note that the output voltage will be reduced.)

Connect the circuit, setting the function generator to 20 V_pp sine wave at a frequency of 1 Hz. (If you cannot obtain a 20 V_pp signal, use the largest signal you can obtain from your generator.) Observe the operation of the circuit; then try speeding up the generator. Look at the waveform across the load resistor with your oscilloscope using the two-channel-difference method. What is the dc voltage across the load resistor? What happens to the output as the generator is speeded up? Try a smaller load resistor. Can you explain your observations?

V_s = 20 V_pp ¦ = 1 Hz     A   R_L  100 kW    B      C₁  100 mF     C₂    100 mF  

Figure 20-4

Report for Experiment 20

ABSTRACT:

DATA:

Table 20-1

Capacitor	Listed Value	Ohmmeter Test Pass/Fail	Voltmeter Test Pass/Fail
C1	100 mF
C2	47 mF
C3	1.0 mF
C4	0.1 mF
C5	0.01 mF

 

                                                         Table 20-2

Step	V1	V2	Q1	Q2	Observations
4
5
6
7
8
9
10

 

22 Inductors

OBJECTIVES:

After performing this experiment, you will be able to:

1. Describe the effect of Lenz’s law in a circuit.

2. Measure the time constant of an LR circuit and test the effect of series and parallel inductances on the time constant.

READING:

Floyd, Principles of Electric Circuits, Sections 14-1 through 14—5

MATERIALS NEEDED:

Two 7 H inductors (approximate value) (Triad C-8X or equivalent) (second inductor may be shared with another experimenter)

One neon bulb (NE-2 or equivalent)

One 33 kW) resistor

For Further Investigation: One unknown inductor

Application Problem: One 100 F capacitor, one 1N4001 diode

SUMMARY OF THEORY:

When a current flows through a coil of wire, a magnetic field is created in the region surrounding the wire. This electromagnetic field accompanies any moving electric charge and 1s proportional to the magnitude of the current. If the current in the coil changes, the electromagnetic field causes a voltage to be induced across the coil which opposes the change. This property, which causes a voltage to oppose a change in current, is called inductance.

Inductance is the electrical equivalent of inertia in a mechanical system. It opposes a change in current in a manner similar to how capacitance opposed a change in voltage. This property of inductance is described by Lenz’s law. According to Lenz’s law, an inductor develops a voltage across it which counters the effect of a change in current in the circuit. Inductance is measured in henries. One henry is defined as the quantity of inductance present when one volt is generated as a result of a current changing at the rate of one ampere per second. Coils made to provide a specific amount of inductance are called inductors.

When inductors are connected in series, the total inductance is the sum of the individual inductors. This is similar to resistors connected in series. Likewise, the formula for parallel inductors is similar to the formula for parallel resistors. Unlike resistors, an additional effect can appear in inductive circuits. This effect is called mutual inductance and is caused by interaction of the magnetic fields. The total inductance can be either increased or decreased due to mutual inductance.

Inductive circuits have a time constant associated with them, just as capacitive circuits do, except the rising exponential curve is a picture of the current in the circuit rather than the voltage, as in the case of the capacitive circuit. Unlike the capacitive circuit, if the resistance is greater, the time constant is shorter.

The time constant is found from the equation:

                      t = L/R

Where t = time constant in seconds,

L = inductance in henries and

R = resistance in ohms. |

PROCEDURE:

1. In this step, you can observe the effect of Lenz’s law. Connect the circuit shown in Figure 22-1 with a neon bulb in parallel with a large inductor. Neon bulbs contain two insulated electrodes in a glass envelope containing neon gas. The gas will not conduct unless the voltage reaches approximately 70 V. When the gas conducts, the bulb will glow. When the switch is closed, dc current in the inductor is determined by the inductor’s winding resistance. Close and open S₁ several times and describe your observations in the report.

2. Find out if the neon bulb will fire if the voltage is lowered. How low can you reduce the voltage source and still observe the bulb? Record your observations in the report.

+12 V    S₁    Neon bulb    L₁  7H

Figure 22-1

Function generation ¦ = 300 Hz    V_S = 1.0 V_pp Square wave

R₁   33 kW      L₁     7H

Figure 22-2

3. Connect the circuit shown in Figure 22—2. This circuit will be used to view the waveforms from a square-wave generator. Set the generator, V_S, for a 1.0 V_pp square wave at a frequency of 300 Hz. This frequency is chosen to allow sufficient time to see the effects of the time constant. View the generator voltage on CHI of a two-channel oscilloscope and the inductor waveform on CH2. If both channels are calibrated and have the VOLTS/DIV controls set to the same setting, you will be able to see the voltage across the resistor using the difference channel. Set the oscilloscope SEC/DIV control to 0.5 ms/div. Sketch the waveforms you see on Plot 22-1.

4. Compute the time constant for the circuit. Enter the computed value in Table 22-1. Now measure the time constant by viewing the waveform across the resistor. The resistor voltage has the same shape as the current in the circuit, so you can measure the time constant by finding the time required for the resistor voltage to change from 0 to 63% of its final value.* Stretch the waveform

* Alternatively, you can measure the rise time and convert the reading to time constant. The relation between rise time and time constant is t = t_r/2.20. Across the oscilloscope screen to make an accurate time measurement. Enter the measured time constant in Table 22-1.

5. When inductors are connected in series, the total inductance increases. When they are connected in parallel, the total inductance decreases. To see the effect of parallel inductors, connect one end of a second 7 H inductor to the first inductor. Then, while observing the waveform across the first inductor, complete the parallel connection of the inductors. You should observe the waveform change as you alternately add or remove the parallel inductor.

You can see the effect of series inductors by placing the two inductors in series. While observing the waveform across both inductors, short out one of the inductors with a jumper wire. Note what happens to the voltage waveforms across the resistor and the inductors. Describe your observations in the space provided in the report.

FOR FURTHER INVESTIGATION:

Suggest a method in which you could use a square-wave from a function generator and a known resistor to determine the inductance of an unknown inductor. Then obtain an unknown inductor from your instructor and measure its inductance. Report on your method, your result, and how your result compares to the accepted value for the inductor.

APPLICATION PROBLEM:

A switching power supply is a very efficient way to produce regulated dc. It uses transistor switches to rapidly “chop” unregulated dc into pulses. The transistors are controlled by circuits that sense the output voltage and vary the duty cycle of the pulses. This action causes the output voltage to be controlled by varying the on time of the pulses. The pulse train is smoothed by inductive and capacitor filters before the load.

The circuit shown in Figure 22-3 represents the output portion of a switching regulator. The pulse generator represents the output from the switching transistors. The inductor is normally a “pot-core” nonsaturable inductor. The inductor in this experiment is much larger than is usually needed; however, it will serve to illustrate the filtering action. When the pulse generator is ON, current flows through the inductor in the direction indicated in Figure 22—3(a). Note that the diode is OFF. When the pulse generator

+ L₁ -  7H     D₁  IN4001 OFF        C₁  100 mF  R_L 33 kW

1. Pulse Generator ON, Diode OFF

+ L₁ -   D₁   IN4001 ON       C₁  100 mF    R_L  33 kW

2. Pulse Generator OFF, Diode ON

Figure 22-3

Report for Experiment 22

ABSTRACT: 

DATA:

Observations from Step 1:

______________________________________________________

Observations from Step 2:

_______________________________________________________

Table 22-1

	Computed Time	Measured Time
Time constant, t

 

V_s = …………………

V_L = ……………..

V_R = ……….

Observation from Stem 5:

_______________________________________