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School Of Engineering Technology and Applied Science (SETAS)
Advanced Manufacturing and Automation Technology (AMAT)
EET 117 – Lab Instructions
Section : __________________
Date: Lab #8
GROUP : ___________________
Series-Parallel Circuits Based on Experiments in Basic Circuits by David Buchla
Name : __________
Name :______________________________
Status and Signature : Objectives: 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.
Required Instruments and Components:
Power supply
DMM (Digital Multi-meter)
Breadboard
Alligator test leads (from the EET-117 labkit)
Resistors: 2.2 kΩ, 4.7 kΩ, 5.6 kΩ, 10.0 kΩ (from the EET-117 labkit)
Procedure
1.
Obtain the resistors listed in the Table 1. Measure each resistor and record the measured value in the table. Reminder of steps to measure resistance using lab DMM (reference to the manual):
1. Connect the device under test to the instrument, as shown:
2. Select a resistance measurement function:
• Press 2 to select 2-wire ohms.
Ω
Table 1. Measured and computed resistance values (use 3 significant digits, metric prefixes).
Component
Listed Value
Measured Value
Marks
R
1
2.2 kΩ
2.193 kOhms
/1
R
2
4.7 kΩ
4.63 kOhms
/1
R
3
5.6 kΩ
5.614 kOhms
/1
R
4
10.0 kΩ
9.969 kOhms
/1
Total:
/4 2.
Connect the circuit shown in Figure 1. Notice that the identical current is through R
1
, and R
4
so we know that they are in series. R
2
has both ends connected directly to R
3
so these resistors are in parallel.
Fig. 1
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 R
2
and R
3
with one equivalent resistor. Label the equivalent resistor R
23
. Draw the equivalent series circuit in the space provided below. Show the value of all components, including R
23.
R(2)(3) = (4.7 kOhms)(5.6 kOhms)/4.7 kOhms + 5.6 kOhms
R(2)(3) = 2.56 kOhms
Marks: / 10
4.
The equivalent circuit you drew above (in step 3) is a series circuit. Compute the total resistance
R
T
, total current I
T
, Voltages
on different parts of this equivalent circuit (
V
1
, V
23
,
V
4
) and enter it in
Table 2. Reference sub-steps:
a)
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. b)
In the equivalent series circuit, the total current is through R
1
, R
23
, and R
4
. c)
The voltage drop across each of these resistors can be found by applying Ohm's law to each
resistor. Compute V
1
, V
23
, and V
4
using this method. d)
Use V
23
and Ohm's law to compute the current in R
2
and R
3
(I
2
and I
3
) of the original circuit.
As a check, verify that the computed sum of I
2
and I
3
is equal to the computed total current.
5.
Measure, enter and compare values in the column beside of Table 2. Important reminder: don’t
forget to disconnect the power supply when measuring the total resistance. When
measuring current always connect ammeter in series and select appropriate connections
on DMM. Table 2. Measured and computed values (use 3 significant digits, metric prefixes).
Computed (Ohm’s Law)
Measured
Marks
V
s
12.0 V
11.962 V
/2
R
T
14.76
14.58
/4
I
T
0.813 mA
0.829 mA
/4
V
1
1.79V
1.81V
/4
V
23
2.08V
2.1V
/4
V
4
8.13V
8.19V
/4
I
2
0.44 mA
/2
I
3
0.37 mA
/2
Total:
/26
6.
The voltage divider rule can be applied directly to the equivalent series circuit to find the voltages
across R
1
, R
23
, and R
4
. Find V
1
, V 23
, and V
4
using the voltage divider rule. Tabulate the results in
Table 2 and place results in Table 3.
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Table 3. Measured and computed values (use 3 significant digits, metric prefixes).
COMPUTED (Voltage Divider Formula: show formula and calculations) Mark
s
V
1
[2.2 kOhms(R1)/14.76 kOhms(R-Total)]12V(V-Source) = 1.79V
/2
V
23
(2.56 kOhms/14.76 kOhms)12V = 2.08V
/2
V
4
(10 kOhms/14.76 kOhms)12V = 8.13V
/2
Total:
/6
7.
The current divider rule can be applied to the parallel circuit to find the current in its branches
(e.g. I
2 and I
3
). Tabulate the results from Table 2 and place results in Table 4 showing formulas and
calculations. Table 4. Measured and computed values (use 3 significant digits, metric prefixes).
COMPUTED (Current Divider Formula: show formula and calculations) Mark
s
I
2
I-2 = [R-3/(R-2+R-3)]I-Total = [5.6k/(4.7k + 5.6k)]0.813 mA = 0.442 mA
/2
I
3
I-3 = [R-2/(R-2+R-3)I-Total = [4.7k/(4.7k + 5.6k)]0.813 mA = 0.371 mA
/2
Total:
/4
8.
Change the circuit to the circuit shown in Figure 2. 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 5.
Fig. 2
R-12 = R-1 + R-2 = 6.9 kOhms
R-34 = R-3 + R-4 = 15.6 kOhms
Marks: / 10
Table 5. Measured and computed values (use 3 significant digits, metric prefixes).
Computed
Measured
Marks
R
12
6.9 kOhms
6.84 k
/4
R
34
15.6 kOhms
15.41 k
/4
R
T
4.78k
4.74k
/4
I
T
2.51 mA
2.55 mA
/4
I
12
1.74 mA
1.78 mA
/4
I
34
0.77 mA
0.78 mA
/4
V
1
3.84V
3.84V
/4
V
2
8.19V
8.19V
/4
V
3
4.33V
4.33V
/4
V
4
7.7V
7.7V
/4
Total:
/40
REVIEW QUESTIONS
1.
The voltage divider rule was developed for a series circuit, yet it was applied to the circuit in
Figure 1.
(a) Explain.
That’s because we can turn parallel circuits into series circuits.
Marks: / 2
(b) Could the voltage divider rule be applied to the circuit in Figure 2? Explain your answer.
Yes it can, Simply find R-equivalent by dividing the product by the sum of the total resistance in each branch. This will create a series circuit which will allow you to apply the voltage divider rule using Ohms law.
Marks: / 2
2.
As a check on your solution of the circuit in Figure 2, apply Kirchhoff's voltage law to each of two
separate paths around the circuit. Show the application of the law.
V-Source = 12, V-12 = 3.84V + 8.19V = 12.03V, V-34 = 4.33V + 7.7V = 12.03
V-Source – V-12 = -0.03V, V-source – V-34 = -0.03V
Marks: / 4
3.
Show the application of Kirchhoff's current law to the junction of R
2
and R
4
of the circuit in Figure 2.
I-Total – (I-12 + I-34) = 2.51mA – (1.74mA + 0.77mA) = 2.51mA – 2.51mA = 0A
Marks: / 4
4.
In the circuit of Figure 2, assume you found that I
T
was the same as the current in R
3
and R
4
.
(a)
What are the possible problems?
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Marks: / 4
(b) How would you isolate the specific problem?
Marks: / 2
5.
The circuit in Figure 3 has three equal resistors.
Fig. 3
(a) If the voltmeter reads +5.0 V, find the voltage drop across R
1 –> V
1
.
R-1 = 10V
Marks: / 4
(b) What is the source voltage (Vs)?
15V
Marks: / 4
6.
What basic rules determine if two resistors in a series-parallel combination circuit are connected
in series or in parallel?
Marks: / 4
Conclusions. The conclusion summarizes the important points of the laboratory work. You must
analyze the examples to add emphasis to significant points. You must also include features and/or
things you have done /benefits of a particular procedure, instrument, component, or circuit directly
related to the experiment
.
This Lab taught me how to determine resistance, voltage and current of both series and parallel
circuits. We discussed how to find voltage drops and currents for each resistor on the circuit board,
along with how to find equivalent resistance in parallel circuits (both how to calculate and measure
each unit).
Marks: / 20
Rubric-Grading
Criteria
Max.
Marks
Punctuality
10
Lab Safety
20
Procedure
100
Review Questions
30
Conclusion
20
Neatness, Spelling, Grammar, and Sentence Structure
10
Total:
/190
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