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UMass Boston Physics 182 Fall 2021
Kirchhoff’s Rules in Series and Parallel Circuits
Introduction
As we learned from the previous experiment (Ohm’s law), an electrical circuit is any continuous path or
array of paths along which current may flow. A circuit usually contains a battery or other sources of EMF
(electromotive force) to create the current. Without a source of energy to drive the circuit no current
(electric charge) will flow. Between the terminals of our power source can be any combination of
elements through which the electrons may pass; anything from a single wire to a complicated collection
of wires, diodes, transistors, and other circuit elements as resistors and capacitors.
Whatever the elements that make up the circuit there are some simple rules that must be obeyed. Two of
these rules are Kirchhoff’s laws regarding current (flow of electric charge) and voltage (electrical
potential difference). To explore these concepts, we will construct circuits that combine resistors in
different arrangements (series and in parallel), and then measure current and voltages at various points in
these circuits. We will compare these values to theoretically predicted values to validate Kirchhoff’s
Laws.
Since, as far as we know, charge can be neither created nor destroyed, if we pick a single point in a circuit
all of the charge that flows into that point must also flow out of it. Derived from the conservation of
charge, and put in terms of currents Kirchhoff’s law, this means that at a given node point (see diagram,
below) the sum of the currents flowing into and out of that node will be equal. This can be written as
|𝐼!”| = |𝐼#\$%| and in Figure 1, below, |𝐼&| = |𝐼+ 𝐼(|.
As we saw in our previous experiment, the sign of the current depends on how the current is measured. In
this experiment we can choose the low connection point at the node for all three measurements. This will
yield a positive current measurement for I
1 and a negative current measurement for I2 and I3. Adding these
three current values adds to zero:
𝐼1 + (-𝐼2) + (- 𝐼3 ) = 0. Note in this case one current is measured as a
positive, flowing
into the node, and two of the currents are negative, they flow out of the node, where the
node is the low potential point. The negative (-) black node dot is the low connection reference point and
the three positive (+) dots are the high connection reference points for the voltage measurements when
checking Procedure Part C. Kirchhoff’s Current Law. It should be noted that this node is an arbitrary
reference point for making measurements. It is also possible to measure each current independently,
changing the low reference point for each measurement to get a positive current reading,
both of these
methods are described in the Procedure section C.
Figure 1
UMass Boston Physics 182 Fall 2021
Kirchhoff’s law for voltages in a circuit is a consequence of the fact that the electric field exerts a
conservative force. Thought of in terms of potential energies, the potential energy of a charge at a
particular point in space does not depend on the path the charge took to get to that point. Since the
potential energy of a charge in a circuit is its charge,
𝑄, times the voltage at the point in the circuit, V, the
sum of the voltages measured around any loop in a circuit must add to zero. Otherwise, if the charge went
around that loop its energy would be changed when it returned to its starting point. This is shown
schematically in Figure 2. Here the voltage sum will be zero,
𝑉𝑎 → 𝑏 + 𝑉𝑏 → 𝑐 + 𝑉𝑐 → 𝑑 + 𝑉𝑑 → 𝑎 = 0.
Assuming that the bottom left-hand node “
a” is at the higher potential 𝑉1, 𝑉2 and 𝑉3 would be positive and
𝑉𝑑 → 𝑎 would be negative as the voltages are measured counterclockwise.
In this part of the lab,
𝑉𝑒𝑚𝑓 = 10.00𝑉 will be the power supply that is driving the circuit. Recall from the
previous experiment,
𝑅 = 𝑉/𝐼, therefore 𝑉 = 𝐼𝑅. In a series circuit the current, 𝐼, is a constant. This
yields:
𝑉& + 𝑉+ 𝑉( – 𝑉)*+ = 0
𝑉
& + 𝑉+ 𝑉( = 𝑉)*+
𝐼𝑅& + 𝐼𝑅+ 𝐼𝑅( = 𝑉)*+
Figure 2
Applying Kirchhoff’s Laws to all the loops and all the nodes in a complicated circuit will result in a series
of algebraic equations which can be solved to uniquely determine all of the voltages and currents in our
circuit. This is essential in circuit design.
Experimental Apparatus and Procedure
Apparatus
1. DC power supply.
2. Agilent 34405A digital volt-ohmmeter (to be used for voltage and resistance measurements).
3. Fluke digital multi-meter (to be used for all current measurements).
4. Carbon resistors and connecting cables.
If you are curious, read the manuals concerning the operation of the power supply and the digital meters
or read the short description in the appendix.

UMass Boston Physics 182 Fall 2021
Procedure:
A. Direct Resistance Measurements with a Digital Ohmmeter.
1. Measure the resistances of the carbon resistors and record your three directly measured values: 𝑅1, 𝑑𝑚,
𝑅2
,
𝑑𝑚
, and 𝑅3
,
𝑑𝑚
using the digital voltmeter (Agilent 34405A) set to measure resistance in ohms (use the
button labeled with Ω).
2. Connect the resistors in parallel and series and measure and record the equivalent parallel and series
resistance of these networks,
𝑅𝑝, 𝑑𝑚, and 𝑅𝑠, 𝑑𝑚.
Remember: When measuring resistance of a component or network, the network should
not be connected
to any other circuitry.
Figure 3
B. Kirchhoff’s Voltage Law
Assemble the series circuit using three carbon resistors as shown in the Figure 4. Connect the power
supply across this network with the current meter (Fluke multimeter) in series. The current meter
measures the current (1 mA = 0.001 A, where A = ampere). The current is a constant in this series circuit.
Set the power supply to approximately V
emf = 10V.
Do not record the number on the power supply display, this display is less precise than the Agilent
multimeter. Use the Agilent multimeter to precisely measure the power supply voltage.
Record the number you set as it shows on the Agilent, (eg Vemf = 10.0076 V.)
Record the current that is being directly measured by the Fluke current meter as Idm.
Directly measure the voltage drops across each resistor using the voltmeter and record the three
directly measured values as
V1,dm, V2,dm, and V3,dm.
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Figure 4
C. Kirchhoff’s Current Law
In this section we will be studying Kirchhoff’s current law. The circuit we will be examining is shown in
Figure 5. The node we are going to explore is the point where the three resistors connect. Because we
need to insert the current meter in series with each leg of the circuit of interest there will be a series of
measurements with the current meter in different place in the circuit. All measurement will be made with
the same voltage, 5.00 volts, applied to the circuit.
Figure 5
First, record the voltage across each of the three resistors with the low side always at the node. As 𝑅2 and
𝑅3 form a parallel connection, 𝑉2 and 𝑉3 should be the same negative voltage. V1 is a positive voltage.
Next, insert the current meter between the end of the first resistor and the low side which is the node. To
do so, cables will have to be removed when the current meter is connected. Record the current running
through the first resistor. Next, insert the current meter between the second resistor and the node. Record
the current running through the second resistor. Finally, insert the current meter between the third resistor
and the node. Record the current running through the third resistor. Using this procedure both
𝐼2, 𝑑𝑚
and 𝐼3
,
𝑑𝑚
should have a negative reading. 𝐼1, 𝑑𝑚 is a positive current.
There is an alternative method to measure
𝐼&, 𝐼, and 𝐼( which results in a positive value for each. This
requires building the circuit three times, each time placing the current meter in a different location to
capture the appropriate current. The circuit diagrams for this procedure are shown in Figure 6.

UMass Boston Physics 182 Fall 2021
Figure 6
In either case, the absolute value of the currents flowing into the node should be equal to the absolute
value of the currents flowing out of the node:
|𝐼+ 𝐼(| = |𝐼&|.
Calculations and Analysis.
The focus of this experiment is not to measure a single value (such as the latent heat or specific heat of a
material), but rather a series of experimental tests of theoretical predictions in circuits. In each step, a
theoretical prediction is formed and checked against some combination of directly measured quantities.
NOTE: For the Calculations section of your lab report, you must show the calculation for each
check. You do not need to show the percent difference calculations, instead, you should present
all the percent differences in table form. For this experiment percent differences should be
calculated using the directly measured values as the accepted value:
%𝑑𝑖𝑓𝑓 = 100 × |”01#”2342|
01
You also are only required to show error calculations for Check 3 and Check 6. Remember,
these may require multiple steps. You must include a Precision versus Accuracy test for Check 3
and Check 6 as well.
Analysis of Resistances in Parallel and Series
Check 1
First, we assembled three resistors in a parallel network. What does the theory claim? It claims that the
reciprocal of the combined resistance equals the sum of the reciprocals of each resistor. In other words,
1
𝑅
\$,&'(&
=
1𝑅
)
+
1𝑅
*
+
1𝑅
+
Giving us our theoretical calculation,
𝑅
\$,&'(& = +𝑅1) + 𝑅1* + 𝑅1+,#)
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Compare the calculated theoretical value (𝑅\$,,-) to the directly measured value (𝑅\$,./) by calculating
the percent difference. Here, we use
𝑅\$,,- as the accepted value.
%𝑑𝑖𝑓𝑓) = 100 ∙ .𝑅\$,&’𝑅(&\$,./𝑅\$,./.
Check 2
When we have three resistors assembled in a series network, we calculate our theoretically predicted
resistance as the sum of each individual resistor:
𝑅
0,,- = 𝑅) + 𝑅+ + 𝑅+
Check this theoretical claim versus what was directly measured.
%𝑑𝑖𝑓𝑓* = 100 ∙ .𝑅0,&’𝑅(&0,./𝑅0,./.
Analysis of a series circuit and Kirchhoff’s Voltage Law
Check 3
Our theoretical claim is that the current passing through the series network can be calculated using the
quotient of the applied emf and series resistance. Calculate this predicted current using the directly
measured series resistance.
𝐼0
,&'(&
=
𝑉
1/2,./
𝑅
0,./
Check this claim with the directly measured current.
%𝑑𝑖𝑓𝑓+ = 100 ∙ .𝐼0,&'(& 𝐼0,./𝐼0,./.
Check 4
In the series circuit with ~10 volts across it, our theoretical claim is that the voltage drop across each of
the three resistors should sum to the three directly measured voltages.
𝑉
1/2,&'(& = 𝑉),./ + 𝑉*,./ + 𝑉+,./
Check this directly with the measured voltage from the power supply.
UMass Boston Physics 182 Fall 2021
%𝑑𝑖𝑓𝑓3 = 100 ∙ .𝑉1/2,&’ 𝑉1/2 (& ,./ 𝑉1/2,./.
Check 5
Using just the directly measured resistances and current, our next theoretical claim is that those quantities
would be enough to find the voltage drop across each resistor.
𝑉),&'(& = 𝑅),./ ∙ 𝐼0,./
𝑉*,&'(& = 𝑅*,./ ∙ 𝐼0,./
𝑉+,&'(& = 𝑅+,./ ∙ 𝐼0,./
Check each against what was directly measured,
%𝑑𝑖𝑓𝑓4′ = 100 ∙ .𝑉),&'(& 𝑉),./𝑉),./.
%𝑑𝑖𝑓𝑓
45 = 100 ∙ .𝑉*,&'(& 𝑉*,./𝑉*,./.
%𝑑𝑖𝑓𝑓
4& = 100 ∙ .𝑉+,&'(& 𝑉+,./𝑉+,./.
Check 6
Now we are ready to check Kirchhoff’s loop law. The claim is that all the voltage decreases and increases
around any closed loop must sum to zero. Due to this being a series circuit, we have essentially already

UMass Boston Physics 182 Fall 2021
accomplished this with Check 4; however, the loop law is important enough to merit its own check.
Figure 7
We draw our loop in the same direction as the assumed direction of the current. Starting at the bottom left
corner, the loop law claims that
𝑉
1/2 – 𝐼𝑅) – 𝐼𝑅* – 𝐼𝑅+ = 0
Switching to the notations we already used, we check the loop law by seeing how close to zero this
calculation comes out to be:
𝑉
1/2,./ – 𝐼0,./(𝑅),./ + 𝑅*,./ + 𝑅+,./) = 0
This result should be reported to two decimal places. The percent difference calculation is impossible,
since the accepted value is 0.00 V.
Analysis of Kirchhoff Current Law.
Check 7
The last two checks deal with Kirchhoff’s node law. In essence, any junction that has current flowing in,
must have the same current flowing out. First, we zoom in on our node.

UMass Boston Physics 182 Fall 2021
Figure 8
To start our checks, we theoretically predict our three currents based on our three voltages.
𝐼),&'(& =
𝑉
),./
𝑅),./
𝐼*,&'(& =
𝑉
*,./
𝑅*,./
𝐼+,&'(& =
𝑉
+,./
𝑅
+,./
Test each against what was directly measured,
%𝑑𝑖𝑓𝑓6′ = 100 ∙ .𝐼),&'(& 𝐼),./𝐼),./.
%𝑑𝑖𝑓𝑓
65 = 100 ∙ .𝐼*,&'(& 𝐼*,./𝐼*,./.
%𝑑𝑖𝑓𝑓
6& = 100 ∙ .𝐼+,&'(& 𝐼+,./𝐼+,./.
Check 8
Confirm Kirchhoff’s node law. The measurements given in the data template use method 1, and so we use
the following equation
𝐼&,-* + 𝐼‘,-* + 𝐼(,-* = 0?
UMass Boston Physics 182 Fall 2021
By keeping the low end of our current meter at the node (method 1) we gain the ability to quickly
measure both the magnitude and the
direction of the current flow. Our assumption is that the current
𝐼& flows into the node, and the currents 𝐼and 𝐼( flow out. This can be confirmed by method one if the
current meter shows
𝐼& as positive and both 𝐼and 𝐼( as negative.
If method 2 were used, we would use the expression:
|𝐼+ 𝐼(| – |𝐼&| = 0. Here we again assume that the
current
𝐼& flows into the node, and the currents 𝐼and 𝐼( flow out. When we connect the current meter in
three different places in the circuit, we expect three positive values. (What would it mean if
𝐼was a
negative value?)
This result should be reported to two decimal places. The percent difference calculation is impossible,
since the accepted value is 0.00 mA.
Reminder:
In this experiment, you do not need to do the error propagation or precision and accuracy comparison for
all 8 checks of Kirchhoff’s laws that were conducted. Instead, only the error calculations and precision
versus accuracy checks are necessary for
Check 3 and Check 6. Remember, it may be necessary to
calculate error in multiple steps.
4. Questions
1. (2 pts each) Let 𝑅𝐴 = 99,000 𝛺 and 𝑅𝐵 = 0.99000 𝛺.
a. Find the equivalent series resistance,
𝑅1, of the two resistors 𝑅2 and 𝑅3.
b. Suggest an
approximation that can be made when two resistors with a large difference in
resistance are combined in series. (“The
approximate resistance of resistors 𝑅2 and 𝑅3in
series is…”)
c. Find the equivalent parallel resistance,
𝑅4, of the two resistors 𝑅2 and 𝑅3.
d. Suggest an
approximation that can be made when two resistors with a large difference in
resistance are combined in parallel.
(This is a useful thing to know in circuit design!)
2. In the circuit used to verify Kirchhoff’s current law, R2 and R3 are connected in parallel (See
Figure 5).
a. (3 pts) Calculate the equivalent parallel resistance
𝑅𝑝 using your measured values of 𝑅2
and 𝑅3.
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b. (3 pts) Your value of 𝑅𝑝, calculated in part (a), is in series with 𝑅1. Calculate 𝑅𝑠 using the
measured value of
𝑅1 and your calculated value of 𝑅𝑝 in part (a).
c. (3 pts) Using the value of
𝑅𝑠 from part (b) above, and 𝑉)*+,-*, calculate the current
through the circuit. Call this current
𝐼𝑐𝑎𝑙.
d. (3 pts) Use percent difference to show the difference between the calculated current in (c)
and the positive sum of
𝐼‘,-* and 𝐼(,-* , (i.e. let 𝐼‘5( = |𝐼+ 𝐼(| ). Use 𝐼678 from part (c)
as the accepted value when doing percent difference.
(Example. If
𝐼‘,-* = – 0.45 𝑚𝐴 and 𝐼(,-* = – 0.93 𝑚𝐴, then 𝐼‘5( = 1.38 𝑚𝐴.)
e. (5 pts) Explain, in your own words, why the positive sum of
𝐼‘,)94 and 𝐼(,)94 should equal
the current in (c).

UMass Boston Physics 182 Fall 2021
Data Sheet. Note: For this experiment an electronic data template is also
provided on Blackboard
Directly measure the three individual resistors. Remember when using the Agilent 34405A for each
measurement in this experiment to adjust the display (with the up and down arrows) to determine the
most precise measurement.

 R1 = ______________ Series Combination Directly measured: Parallel Combination. Directly measured: R2 = ______________ R3 = ______________ Rs,dm = ______________ Rp,dm = ______________

Kirchhoff’s Voltage Law
Applied voltage (~10 volts) across the series network
V
emf = ______________ (Enter as a positive value.)
Current (in mA) through the series directly measured using the Fluke 75.
I
s,dm = ______________
Directly measured voltage drops across the individual resistors. Be mindful of what is the high end of
each resistor. All values should be positive.
V
1,dm = ____________ V2,dm = ____________ V3,dm = ____________
Kirchhoff’s Current Law
Applied voltage (~5 volts) across the network
V
emf = ______________ (Enter as a positive value.)
Keeping the low end of the voltmeter at the node, directly measure the voltage across the resistors.
Results should be both positive and negative.

 V1 = ____________ V2 = ____________ V3 = ____________ Keeping the low end of the current meter at the node, directly measure the current going in or out of the node. Results should be both positive and negative. I1,dm = ____________ I2,dm = ____________ I3,dm = ____________

UMass Boston Physics 182 Fall 2021
Appendix: Instrumentation
1. The DC Voltmeter measures the voltage difference between two points to which its terminals are
connected. The voltmeter is always connected
in parallel to the part of the circuit across which the
potential difference is to be measured (Figure A1). In this experiment the Agilent 34405A multi-meter
(multimeter) is used to measure voltage differences and the resistance of the carbon resistor.
Figure 9
2. The DC Milliammeter current meter measures the electric current between any two points to which its
terminals are connected. To measure the current in any part of the circuit, the circuit must be broken at that
point and the current meter must be inserted in the gap with loose ends connected to its terminals, i.e., the
meter is connected
in series with that part of the circuit where the current is to be measured (Figure A2).
The Fluke 75 battery powered multi-meter is used to measure the current in milliamperes in this
experiment.
3.
Figure 10
4. Connecting wire leads. Assumed to be made of a good conducting material, e.g., a metal such as copper,
and of a large enough cross-section to have negligible resistance and voltage drop across them.
5. The
DC Power supply is a device used to supply a constant source of EMF between its output terminals.
The electric current flows internally in the power supply from the minus to the plus terminal. In the external
circuit it flows from the plus to the minus terminal (Figure A3). The voltmeter on the actual power supply
is not accurate and should not be used to determine the output voltage. The Agilent 34405A multi-meter is
used to measure voltages on the power supply.