Parallel Circuits
Objectives:
1. Demonstrate that the total resistance in a parallel circuit
decreases as resistors are added.
2. Compute and measure resistance and currents in parallel circuits.
3. Explain how to troubleshoot parallel circuits.
Summary of Theory:
a parallel circuit is one in which there is more than one path for
current to flow, it can be thought of as two parallel lines,
representing conductors, with a voltage source and components
connected between the lines. This idea is illustrated in figure 9-1.The
source voltage appears across each component. Each path for
current is called a branch. The current in any branch is dependent
only on the resistance of that branch and the source voltage.
As more branches are added to a parallel circuit, the total resistance
decreases. If the total current in a circuit increases, with no change
in source voltage, the total resistance must decrease according to
ohm’s law. When connected in parallel the voltage is the same
across all branches but the current is divided at branches.
E = V1 = V2 =V3
1/RT = 1/R1 + 1/R2 + 1/R3
1/RT = R2R3/R1R2R3 + R1R3/R1R2R3 + R1R2/R1R2R3
RT = R1R2R3/ ((R2+R3) + (R1R3) + (R1R2))
If R1 = R2 = R3.
RT = R3/3R2 = R/3
In Parallel: RT< Rmin
Kirchhoff’s current law: The sum of the currents entering a circuit
junction is equal to the sum of the currents leaving the junction. The
current leaving the source must be equal to the sum of the individual
branch currents. While Kirchhoff’s voltage law is developed in the
study of series circuits and the current law is developed in the study
of parallel circuits, both laws are applicable to any circuit.
In parallel circuit, it is the current that is divided between the
resistances. Keep in mind that the larger the resistance, the smaller
the current. The general current divider rule.
In parallel:
IT = I1 + I2
RT = R1R2/ (R1+R2)
V1 = V2 = E
I1 = IT – I2
I1 = V1/R1 = E/R1 = ITRT/R1 = IT (R1R2/R1+R2)/R1 = ITR2/R1+R2
I2 = ITR1/R1+R2
So Ii = ITRj/ Ri+Rj (This is the equation of current divider rule).
Materials Needed:
Resistors: one 2.2kΩ, one 2.7kΩ, one 1kΩ, one 0.2kΩ.
One dc ammeter, 0-10 mA.
Procedure:
1. Obtain the resistors listed in Table 9-1. Measure and record the
value of each resistor.
Table 9-1
Component
R1 =
R2 =
R3 =
R4 =
Listed Value
2.2kΩ
2.7kΩ
1kΩ
0.2kΩ
Measured Value
2.19kΩ
2.684kΩ
1.002kΩ
0.219kΩ
2. In Table 9-2 you will tabulate the total resistance as resistors are
added in parallel. Enter the measured value of R1 in the table.
Then connect R2 in parallel with R1 and measure the total
resistance as shown in figure 9-3. Enter the measured resistance
of R1 in parallel with R2 in Table 9-2.
Table 9-2
R1
RT(measured) 2.19kΩ
IT(measured) 5.45mA
R1 ‖ R 2
1.205kΩ
9.91mA
R1 ‖ R2 ‖ R3
0.547kΩ
21.88mA
R1‖R2‖R3‖R4
0.156kΩ
77.1mA
3. Add R3 in parallel with R1 and R2. Measure the parallel resistance
of all three resistors. Then add R4 in parallel with the other three
resistors and repeat the measurement. Record your results in
Table 9-2.
4. Complete the parallel circuits by adding the voltage source and
the ammeter as shown in figure 9-4. Measure the total current in
the circuit and record it in Table 9-2.
5. Measure the voltage across each resistor. How does the voltage
across each resistor compare to the source voltage?
6. Use ohm’s law to compute the branch current in each resistor.
Use the source voltage and the measured resistances. Tabulate
the computed currents in Table 9-3.
Table 9-3
I1 = Vs/R1
I (computed) 5.479 mA
I2 = Vs/R2
4.470 mA
I3 = Vs/R3
11.976 mA
I4 = Vs/R4
54.794
7. Use the general current divider rule to compute the current in
each branch. Use the total current and total resistance that you
recorded in Table 9-2. Compare the calculation using the current
divider rule with the results using ohm’s law. Show your results in
Table 9-4.
Table 9-4
I1 = RTIT/R1
I (computed) 5.45 mA
The same result.
I2 = RTIT/R2
4.449 mA
I3 = RTIT/R3
11.968 mA
I4 = RTIT/R4
54.920 mA
8. Simulate a burned-out resistor by removing R4 from the circuit.
What is the new total current?
IT = 21.88 mA
Conclusion:
 When connect the circuit in parallel the voltage is the same across
all branches but the current is divided at branches.
 Total resistance less than the smallest resistance in the circuit.
10 Series- Parallel
Combination Circuits
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.
Summary of theory:
Many circuits can be analyzed by applying the ideas developed for
series and parallel circuits to them. 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. The
components that are in series or parallel may be replaced with an
equivalent component. For example, in figure 10-1 (a) we see that
the identical current must flow through 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.
Materials Needed:
Resistors: one 2.2kΩ, one 2.7kΩ, one 1kΩ, one 0.2kΩ.
Procedure:
1. Measure and record the actual values of the four resistors listed
in Table 10-1.
Table 10-1
Component
R1 =
R2 =
R3 =
R4 =
Listed Value
2.2kΩ
2.7kΩ
1kΩ
0.2kΩ
Measured Value
2.19kΩ
2.684kΩ
1.002kΩ
0.219kΩ
2. Connect the circuit shown in figures 10-2. Then answer the
following questions.
(a) Are there any resistors for which the identical current will flow
through the resistors? Answer yes or no for each resistors:
R1 Yes, R2 No , R3 No, R4 Yes .
(b) Does any resistor have both ends directly to both ends of
another resistor? Answer yes or no for each resistors:
R1 Yes, R2 Yes , R3 Yes, R4 Yes .
3. 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.
Table 10-2
RT
IT
V1
V2,3
V4
I2
I3
VT
Computed
Voltage Divider Ohm’s
Law
3.137KΩ
3.137KΩ
3.825mA
8.377 V
6.378 V
3.8 V
1.033
0.765 V
Measured
12.0 V
11.974 V
12.0 V
3.137kΩ
8.350 V
2.787 V
0.839 V