LBYEC11
Electric Circuits 1 Laboratory
Experiment No. 10
“ KIRCHHOFF’S LAWS AND CLOSED LOOPS”
Section/Group No.: EV2/ GROUP2
Group Leader: Laura Dadios
Group Members:
1. Odessah Tomampoc
2. Jason Guillermo
3. Camille Rico
Grade: _____________
Date Performed: 11-19-09
Date Submitted: 11-26-09
Professor: Mr. Vivencio C. Hilario
Instructor’s Signature: ______________
Remarks:
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________________________
EXPERIMENT 10
KIRCHHOFF’S LAWS AND CLOSED LOOPS
I.
OBJECTIVE:

II.
To examine one of the techniques used in electric
analysis which may be applied to closed electric
circuits in which there are one or more sources of
energy.
THEORETICAL CONSIDERATIONS:
Consider the circuit shown in Figure 10.1. In this
case there are several closed
loops. For example, the
loop starting at the positive side of the Ea source and
passing
through EA, R3, EB and R1 may be similarly, the
loop need not contain a source to be
considered
closed
loop.
VOLTAGE LAW
One of the popular techniques used in the analysis of
such loop circuits
employs a law commonly referred to
as Kirchhoff’s Voltage Law. This law is named after the
scientist who first recognized it and can be stated as:
The algebraic sum of the source voltages around any
closed loop is equal to the algebraic sum of the voltage
drops across the resistances in that same closed loop.
This
describes
return to
R3 and R1.
law may be translated into an equation which
the operation of
any loop. For example, let us
the loop in Figure 10.1 which contains EA, EB,
Kirchhoff’s laws ensure that:
EA – EB = E3 + E1 ----------------------------- (Equation 1a)
Where E3 and E1 are the voltage drops across R3 and R1
respectively. Similarly, the equation which describes the
loop involving EB, R3, EA and R2 is:
EA – EB = E3 + E2 ----------------------------- (Equation 1b)
The small R1-R2 loop contains no source, and we have:
0 = E1 – E2 ---------------------------------- (Equation 1c)
One fairly easy method of arriving at the loop
equation for a particular loop is to select arbitrarily a
direction of current flow through one of the sources and
then mark the polarity of each drop such that they are all
compatible with the original choice of current direction.
R1
R2
+
I
To
power
supply
EA
25VDC
-
12V battery
R3
-
+
Figure 10.1
A Circuit with Several Closed Loops
CURRENT LAW
In a previous experiment, you had verified that the
total current I1 in a circuit containing
resistors
in
parallel is equal to the sum of the currents in each of the
branches. This was one demonstration of Kirchhoff’s Current
Law limited to a parallel
network. The law is perfectly
general, however, and is applicable to any
circuit.
It
states that the current entering any junction of an
electric circuit is equal to the current leaving that j
unction.
Applying this law to junction A of Figure 10.2
I1 + I2 – I3 – I4 – I5 = 0
The current law of Kirchhoff may also be stated as
“The sum of the currents
entering a junction is equal
to the sum of the currents leaving the same junction.” As
applied to junction A of Figure 10.2,
I1 + I2 = I3 + I4 + I5
SIGN CONVENTION
For Kirchhoff’s Current Law:


the current is positive in sign if it is entering the
junction
the current is negative in sign if it is leaving the
junction
For Kirchhoff’s Voltage Law:

For sources of EMFs

E is the positive in sign if it is traced from E its
negative to positive terminals;
while

E is the negative in sign if it is traced from E its
positive to negative terminals.

For IR drops

IR is the positive in sign if the resistor is traced
in direction opposite to the flow
of
I
through
the
resistor; while

IR is the negative in sign if the resistor is traced
in direction the same as the flow
of
I
through
the
resistor.
I3
I1
I4
A
I2
I5
Figure 10.2
Current Entering and Leaving Junction A
III. MATERIALS AND EQUIPMENT:
1
1
1
2
1
10
-------------------------------------------------------
EMS POWER SUPPLY MODULE
EMS DC VOLTMETER MODULE
DIGITAL AMMETER (RANGE 10A)
EMS RESISTANCES MODULE (A AND B)
12 – VOLT STORAGE BATTERY
CONNECTING WIRES
IV.
PROCEDURE
1.
Connect the circuit as shown in Figure 10.3.
I1
R1=100Ω
+
R2=75Ω
To
Power
Supply
Module
I2
V
R3=
50Ω
EA=25V
V
-
R4=
100Ω
I4
R5=
75Ω
I5
I3
+
-
EB = 12V
Figure 10.3
The Experimental Circuit
2.
3.
4.
5.
6.
7.
8.
9.
Turn ON the switch of the power supply module.
Set the two sources, EA and EB, to 25 volts and 12
volts, respectively.
Measure and record each of the five voltages E1
across R1, E2 across R2, etc. In Table 10.1
Write the loop equations of the six closed loops in
Figure 10.3.
Substitute the values measured and source values
into each equation and illustrate that each equation
is correct.
Write again the loop equations of the six closed
loops in Figure 10.3. This time in terms of the
branch currents of the circuit.
Using the equations formed in Step 6 together with
the application of Kirchhoff’s Current Law, solve
for the branch currents of the circuit.
Using the result of Step 7, compute and record each
of the five voltages drops in Table 10.1.
Compute for the percent difference between the
measured and computed values of the five voltage
drops and record in Table 10.1.
PRELIMINARY REPORT
DATE PERFORMED: ________________
EXPT. NO. 10
EXPT. TITLE:
KIRCHHOFF’S LAWS AND CLOSED LOOPS
GROUP NO.: ___________
GROUP LEADER:
SIGNATURE:
____________________
______________
GROUP MEMBERS PRESENT: ____________________
____________________
____________________
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I – IV
VVI VII VIII -
:
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:
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TOTAL
: _________________
REMARKS: ____________________________________________________
INSTRUCTOR’S SIGNATURE:________________
GRADE:____________
V.
DATA AND RESULTS:
EA = _________
EB = _________
Table 10.1
Value
Measured
Computed
% Difference
E1
E2
E3
E4
E5
Step 4
Loop
Loop
Loop
Loop
Loop
Loop
Step 5
equation
equation
equation
equation
equation
equation
(1):
(2):
(3):
(4):
(5):
(6):
____________________________
____________________________
____________________________
____________________________
____________________________
____________________________
Loop
Loop
Loop
Loop
Loop
Loop
equation
equation
equation
equation
equation
equation
(1):
(2):
(3):
(4):
(5):
(6):
____________________________
____________________________
____________________________
____________________________
____________________________
____________________________
equation
equation
equation
equation
equation
equation
(1):
(2):
(3):
(4):
(5):
(6):
____________________________
____________________________
____________________________
____________________________
____________________________
____________________________
Step 6
Loop
Loop
Loop
Loop
Loop
Loop
Step 7
Branch Currents:
IT = ____________________________
I1 = ____________________________
I2 = ____________________________
I3 = ____________________________
I4 = ____________________________
I5 = ____________________________
VI.
COMPUTATIONS/GRAPHS:
VII. ANALYSIS AND CONCLUSIONS:
VIII.
QUESTIONS TO BE ANSWERED:
1.
Write loop equations for each of the circuits shown in
Figure 10.4.
I1
R1
R2
R3
-
+
EA
-
EC
R4
-
+
I
EA
I2
+
R1
EB
-
R4
Figure 10.4a
Figure 10.4b
I3 R3
EB
I3
R2
+
R3
R1
I1
+
Figure 10.4c
R2
I2
2.
Substitute the following values into the equations
arrived at no. 1.
EA = 5 volts
R1 = 10K ohms
R4 = 2.7K ohms
3.
EB = 15 volts
R2 = 4.7K ohms
R5 = 15K ohms
Draw a circuit which can be
equation, 6 – 4I – 2I + 3 = 0.
EC = 9 volts
R3 = 20K ohms
represented
by
the