ELP LAB1 REPORT:
KVL AND KCL
BY RAGHAV SINGLA (2023CH10900, GROUP 10, TABLE 6)
AUGUST 17, 2023
Aim:
To verify KVL, KCL, measure input impedance. To study
the effect of some modifications in special circuits.
Apparatus/Material required:
2 power sources(5V), connecting wires, multimeter,
breadboard, resistances ( 2 -100ohm , 2 - 220 ohm , 147 ohm , 1 -150 ohm , 1 – 470 ohm , 5 – 120 ohm , 3150 ohm)
Theory:
Ohm’s law:
it states the relationship between electric current and
potential difference. The current that flows through
most conductors is directly proportional to the voltage
applied to it.
I = V/R
Where I is the current flowing through the conductor,
V is the potential difference applied across the
conductor and R is the Resistance of the conductor.
KVL:
It states that for any closed loop in the circuit, the sum
of potential difference across all the components is
zero.
KCL:
It states that the algebraic sum of all currents entering
and exiting a node must be zero.
The superposition theorem:
It states that a circuit with multiple voltage and current
sources is equal to the sum of simplified circuits using
just one of the sources.
Multimeter:
A measuring instrument that can measure multiple
electrical properties.
Breadboard:
It is a construction base used to build a semipermanent prototype of electronic circuits.
Tellengen’s theorem:
It states that the summation of power delivered is zero
for each branch of any electrical network at any instant
of time. It is mainly applicable for designing the filters
in signal processing.
Part 1 –
Procedure/Setup:
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We connect the network below where V1 = 5
volts; R1 = 100Ω; R2 = 220Ω; R3 = 100Ω; R4 =
220Ω; R5 = 47Ω and R6 = 50Ω.
Then we have to measure each resistance by
opening them out of the circuit and using a
multimeter.
After that, we must measure the voltages across
all resistances in the circuit using a multimeter
and calculate the currents expected to flow
through these resistors.
1.We have to verify KCL for node-A and node-B.
2.We have to verify KVL for all possible loops.
3.We will connect a 470ΩΩ resistor between S
and G; measure the change in the current I5.
4. Now we will connect the same 470Ω resistor
across R6 and measure the change in the
current I5.
Observations:
RESISTANCE
R1
VALUE(OHM)
100 +- 5%
(gold band)
220 +- 5%
100 +- 5%
220 +- 5%
47 +- 5%
150 +- 5%
R2
R3
R4
R5
R6
V DROP
2.478
2.613
1.286
1.326
0.313
1.012
After connecting the 470-ohm resistor between S
and G, the change in I5 = 0.
After connecting the 470-ohm resistor parallel to
R6, the change in I5 = 0.001 A.
Calculations
CURRENT
(I IN AMPERES)
I1 = V1/R1
I2 = V2/R2
I3 = V3/R3
I4 = V4/R4
I5 = V5/R5
I6 = V6/R6
= 2.478/100
= 2.613/220
= 1.286/100
= 1.326/220
= 0.313/47
= 1.012/150
=0.025
=0.012
=0.013
=0.006
=0.007
=0.007
VERIFICATION of KCL for NODE A –
The current entering the node A is I1 and leaving
the node A are I2 and I3. By KCL I1 must be equal
to the sum of I2 and I3.
[I1 = 0.025A, I2= 0.012A, I3=0.013A
I2 + I3 = 0.012A + 0.013A = 0.025A = I1]
Hence KCL is verified for NODE A.
VERIFICATION of KCL for NODE BSimilarly for node B, I3 must be equal to the sum
of I4 and I5.
I3= 0.013, I4=0.006, I5 = 0.007
0.013 = 0.06 + 0.07
Hence KCL has been verified for node B.
VERIFICATION OF KVL FOR ALL POSSIBLE
LOOPS –
the potential difference across all the resistors has
already been calculated above which is
represented by Vn. The sign of the potential
difference would be taken depending upon the
direction of the current
1)LOOP with Power source, R1, R2
V-V1-V2must be 0 which implies,
5 – 2.478- 2.613 must be 0
5 – 2.478 – 2.613 =-0.09 which is nearly 0
2)LOOP with R2, R3, R4
V2– V3 – V4 must be 0 which implies,
2.613 – 1.286 – 1.326 must be 0
2.613 – 1.286 – 1.326 = 0.001 which is nearly 0
3) LOOP with R4, R5, R6
V4 – V5 – V6 must be 0 which implies,
1.326 – 0.313 – 1.012 = 0.001 which is nearly zero
4) LOOP with V, R1, R3, R4
V – V1-V3-V4 must be 0
5 – 2.478 – 1.286 – 1.326 = - 0.09 which is nearly 0
5) LOOP with R2, R3, R5, R6
V2 – V3 – V5 – V6 must be 0.
2.613 – 1.286 – 0.313 – 1.012 = 0.002 which is
nearly 0
6) LOOP with V, R1, R3, R5, R6
V – V1 – V3 – V5 – V6 must be 0
5 – 2.478 – 1.286 – 0.313 – 1.012 = -0.08 which is
nearly 0
After connecting the 470 ohm resistor parallel to
R6, the change in I5 =
(Vnew across R5 – Vold across R5)/R5 – by ohm’s law
= (0.357-0.313)/47 = 0.001 A
Initial circuit setup on the bread board for the first part
Circuit with the 470 ohm resistor (Part 1)
Part-2:
Procedure/Setup:
From the original arrangement 


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We have to move one end of R5 out and
connect a 5 Volt source in series with R5 such
that the positive side of the voltage source
should be towards B. Then measure the voltage
across R2 and calculate the current I2.
We had calculated I2 earlier when there was
only V1 of 5 volts.
Now we have to replace V1 with a short circuit
and again measure the voltage across R2 and
calculate I2.
Then we would be able to verify the
superposition theorem.
Observations –
The superposition principle holds true for the given
circuit.
The current(I2) after the first amendment = 0.015A
The current(I2) after the second amendment =0.003A
Calculations:
After the first amendment:
New voltage across R2 = 3.230
New current through R2(I2) = (3.230)/220 – by ohm’s
law
= 0.015A
After the second amendment:
New voltage across R2= 0.664
New current through R2 = 0.664/220 – by ohm’s law
= 0.003 A
The current calculated in the first part through R2(I2)
=0.012A
By the superposition theorem: The current obtained in
R2 ( that is I2) in the presence of the 2
sources(amendment 1) should be equal to the
algebraic sum of the currents through R2 when the
circuit only contained one of the sources ( original +
amendment 2)
We observe that 0.015A = 0.003A + 0.012A
This verifies the superposition Theorem.
The circuit when the original voltage source is replaced
by a short circuit.
Part-3:
Procedure/Setup:
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We have to setup a circuit as shown below
where each resistor is 120Ω.
Then we will measure the resistance between
nodes A and B with the help of a multimeter.
We will compare the expected resistance
between nodes A and B with the measured
value.
Then we will apply a voltage of 5 volts between
A and B. Then we will measure the voltages
between B-C, B-D, C-D.
Next we will replace C-D with a short-circuit,
measure the voltages across the resistances;
measure the resistances; and calculate the
current in A-A'.
Lastly we will replace C-D with an open-circuit,
once again measure the voltages across the
resistances and calculate the current in A-A'.
Observations:
The measured resistance between A and B = 120 ohm
The expected value of the resistance between A and B
= 120ohm
The expected value matches with the measured value.
After applying a voltage of 5 volts between A and
B–
V(BC)
V(BD)
V(CD)
2.495
2.513
0.010
The voltage across BC approximately = BD = half the
potential difference across AB.
The voltage across CD = 0.010 which approximately = 0
,as if the resistance CD does not exist .
The above 2 observations are made as the given circuit
satisfies a condition known as the balanced
Wheatstone bridge.
In such a case, the galvanometer shows no deflection ,
which means that no current passes through the
central region when the Wheatstone bridge is in a
balanced condition.
After replacing CD with a short circuit:
V(AC)
V(BC)
V(AD)
V(BD)
2.476
2.503
2.478
2.481
The current in entering AA’ must be equal to the sum
of the current leaving A’C and A’D – by KCL
I (AA’) = 0.042 A
Due to balanced Wheatstone condition, no current
flows through the central region. The potential drop
across each resistor except the central one is same due
to the symmetry and simple Kirchoff rules.
After replacing CD with an open circuit:
V(AC)
V(BC)
V(AD)
V(BD)
2.519
2.495
2.502
2.517
The current in entering AA’ must be equal to the sum
of the current leaving A’C and A’D – by KCL
The results observed are almost same as the above
part because the value of R in the centre doesn’t
matter . It may be 0 (open circuit) or several thousand
ohms.
Calculations:
After replacing CD with a short circuit I(A’C) = V(A’C)/R = V(AC)/R = 2.476/120 =0.021
I(A’D)= V(A’D)/R = V(AD)/R = 2.478 /120 = 0.021
I(BC)= V(BC)/R = 2.503/120 = 0.021
I(BD)= V(BD)/R = 2.481/120 = 0.021
Which implies I(AA’)= 0.021A + 0.021A= 0.042A
After replacing CD with an open circuit –
I(A’C) = V(A’C)/R = V(AC)/R = 2.519/120 =0.021
I(A’D)= V(A’D)/R = V(AD)/R = 2.502 /120 = 0.021
I(BC)= V(BC)/R = 2.495/120 = 0.021
I(BD)= V(BD)/R = 2.517/120 = 0.021
Which implies I(AA’)= 0.021A + 0.021A= 0.042A
The original circuit arrangement for the 3rd part
The circuit after the central resistance has been short
circuited.(part 3)
The circuit after the central resistance has been
removed.
Part-4: (mystery)
Procedure/Setup:
We have to consider the circuit of Part-1. We have
already measured all the voltages (and inferred all the
currents) in all the branches of the circuit. Let us call
this set of voltages and currents as {ViA} and {IiA}.
Now we will do a fresh measurement for a similar
circuit, with different values of voltages and
components. The voltage source can be any value of
our choice. The resistors R1 through R6 can be
arbitrarily chosen from 47Ω, 100Ω, 150Ω and 220Ω.
The circuit components shouldn’t be the same as the
previous circuit. We will measure all the voltages, and
infer all the currents; let us call this set of voltages and
currents as {ViB} and {IiB}.
We will first compute ∑i(ViAIiB). That is from the set A,
choose the ith voltage and from set B choose the
ith current. Multiply them and add. We will also
compute ∑i(ViBIiA) and comment on the surprising
result.
Observations :
FOR THE B CIRCUIT R
R1(100 +-5%)
R2(150)
R3(150+-5%)
R4(150+-5%)
R5(100+-5%)
R6(220+-5%)
V(drop)
2.601
2.415
1.432
0.982
0.307
0.672
∑I(ViAIiB) =0
∑I(ViBIiA) =0
We observe these surprising results , which can be well
explained by Tellegen’s theoremwhen we compare the potential drops and currents in
different circuit configurations using Tellegen's
Theorem, we are not only verifying energy
conservation but also validating the foundational
circuit analysis principles that govern the behavior of
electric circuits.
CALCULATIONS:
ViB
IiB = ViB / R
2.601
2.415
1.432
0.982
0.307
0.672
2.601/100 =0.026
2.415/150=0.016
1.432/150=0.010
0.982/150=0.007
0.307/100=0.003
0.672/220=0.003
ViA(Volt)
IiB(Ampere)
(ViAIiB)
Across R1 =2.478
Across R2 =2.613
Across R3 =1.286
Across R4 =1.326
Across R5 =0.313
Across R6 =1.012
Across the
power source =
(-5)
0.026
0.016
0.010
0.007
0.003
0.003
0.026
0.064
0.041
0.013
0.009
0.001
0.003
-0.13
∑I(ViAIiB) = 0.001 which is nearly =0 [measurement
error]
ViB(Volt)
IiA(Ampere)
(VIBIiA)
Across R1 2.601
Across R2 2.415
Across R3 1.432
Across R4-0.982
Across R5 -0.307
Across R6 -0.672
Across the
power source =
(-5)
=0.025
0.065
=0.012
0.029
=0.013
0.019
=0.006
=0.007
=0.007
0.025
0.006
0.002
0.005
-0.125
∑i(VIBIiA)= 0.001 which is approximately =0
[measurement error]
The circuit arrangement for the 4th part.
CONCLUSION
1. KCL and KVL have been verified.
2. Superposition principle holds true.
3. Energy is conserved.(4th part)
SOURCES OF ERROR
1. The error in resistance : All the resistances are gold
band resistances , which means that they will have
approximately 5% error . For example 100ohm+5% = (95 ohm to 105ohm)
2. The error in multimeter :
For DC voltage – 0.5% , For Resistance – 0.9%
(source – user manual )
3. The calculation/measurement error : All the
calculations done by me involved approximation
till the 3rd decimal place
**The calculations where the expected result was 0
but the calculation showed approximately 0 was
because of the above 3 errors in the measurement.
**The voltage across the power sources in the above
parts was taken negative as we have to compute the
sum algebraically.
PRECAUTIONS
1. Properly dress up for the lab (shoes are a must)
2. Don’t operate any electrical circuit with wet
hands.
3. The DC voltage source should have 2 functioning
portals.
4. All the lines in the bread board should be
functional .
5. The multimeter should be working properly.
6. Don’t forget to look at the colour bands of the
resistance to precisely calculate the error.
THANK YOU