Lab E x p e r i m e n t # 08
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Title: Verification of Current Divider Rule.
Objective: Objective of this practical is to understand and verify that current is divided in a
parallel resistive circuit.
Required Apparatus:- Few resistors, bread board, multimeter, DC voltage source,
connecting wire etc.
Theory:- Current divider rule states that “Current is divided in a parallel circuit, whereas sum
of all currents through all resistors is equal to total current”
Procedure:- Connect two resistors “R1” and “R2” in parallel and then connect positive
terminal of DC voltage source “Vs” with terminal “a” of R1 & R2 and negative terminal of
DC voltage source “Vs” with terminal “b” of R1 & R2 as shown in figure 01. Find out
current through R1 and R2 with the help of multimeter as well as formula IR1=(IT x RT)/R1
where
I R1 : Current through resistor R1
R1 : Value of resistor R1
IT : Total current
RT : Total resistance ((R1*R2)/(R1+R2))
Figure 01: Parallel resistive circuit
Observation:- Current through R1 is 100mA and current through R2 is 10mA, whereas
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sum of both currents through R1 and R2 is equal to total current (Vs/RT)=110mA.
Table 01: Verification of Current Divider Rule
S.No
R1
R2
Vs
Figure 01
100 ohm
1K-ohm
10V
RT 
R1R2
R1  R2
90.909 ohm
IT 
Vs
RT
110mA
I R1 
RT IT
R1
100mA
I R2 
RT IT
R2
10mA
I R1  I R 2
110mA
Answer the following questions:Question 01: What is current divider rule?
Question 02: Fill in the following table using formula, multisim or hardware.
S.No
R1
R2
Vs
Example
100 ohm
1K-ohm
10V
1
200 ohm
1K-ohm
10V
2
500 ohm
1K-ohm
10V
3
1K-ohm
1K-ohm
10V
4
2K-ohm
1K-ohm
10V
RT 
R1 R2
R1  R2
90.909 ohm
IT 
Vs
RT
110mA
I R1 
RT IT
R1
100mA
Question 03: Discuss the results obtained in question 02.
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I R2 
RT IT
R2
10mA
I R1  I R 2
110mA
Lab E x p e r i m e n t # 09
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Title: Superposition Theorem.
Objective: Objective of this practical is to learn about Superposition Theorem; and how this
theorem is used to find out voltage at and/or current through some particular resistor in a
complex circuit.
Required Apparatus:- Few resistors, bread board, multimeter, DC voltage sources,
connecting wire etc.
Theory:- In some circuits there can be more than one energy (current and/or voltage)
sources, for such circuits Superposition theorem provides a method for analysis.
This theorem states that “The current in any given branch of a multiple-source circuit can be
found by determining the currents in that particular branch produced by each source acting
alone, with all other sources replaced by their internal resistances. The total current in the
branch is the algebraic sum of the individual currents in that branch”.
Steps for applying superposition theorem:
 Step 01: Leave one voltage (or current) source at a time, replace other sources with their
internal resistances
 Step 02: Determine particular current (or voltage)
 Step 03: Take next source and repeat step 1 & 2
 Step 04: Actual current in particular branch is sum of all individual currents. Voltage can
be found using Ohm’s law then.
Procedure:- Realize the circuit on breadboard as shown in figure 01. Now find out current
through R2 with the help of multimeter and then solve the circuit step by step with the help of
superposition theorem to verify that current found through R2 with both methods is same
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Figure 01: Complex circuit with two energy sources
Current found through R2 with the help of multimeter is 50mA.
Now by applying Superposition theorem current through R2 can be found as follows:
Step 01: Leave source 1 and replace source 2 with its internal resistance as shown in figure
02.
Figure 02: Circuit with energy source 2 replaced with its internal resistance
Step 02: Current found through R2 with help of formulas/multimeter is equal to
I(S1)=33.3mA.
Step 03: Leave source 2 and replace source 1 with its internal resistance as shown in figure
03.
Figure 03: Circuit with energy source 1 replaced with its internal resistance
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Step 04: Current found through R2 with help of formulas/multimeter is equal to
I(S2)=16.7mA.
Step 05: Total current through R2 is the sum of I(S1) and I(S2). Which means
I(S1+S2)=I(S1)+I(S2)=33.3mA+16.7mA=50mA
Observation:- Value of current found through R2 with the help of multimeter and
Superposition Theorem is same, Hence it proved that Superposition Theorem provides a
method for the analysis of a complex circuit containing more than more than one energy
sources.
Answer the following questions:Question 01: When and why Superposition Theorem is used?
Question 02: What does Superposition Theorem state?
Question 03: What are different steps of Superposition Theorem?
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Lab E x p e r i m e n t # 10
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Title: Thevenin’s Theorem.
Objective: Objective of this practical is to learn about Thevenin’s Theorem; and how this
theorem is used to convert a complex resistive DC circuit into thevenin equivalent circuit.
Required Apparatus:- Few resistors, bread board, multimeter, DC voltage sources,
connecting wire etc.
Theory:- Thevenin’s theorem provides a method for simplifying any two terminal complex
resistive DC circuit to an standard equivalent form. Thevenin’s equivalent circuit consists of an
equivalent voltage source and a resistor in series. A complex circuit and its Thevenin
equivalent circuit is shown in figure 01.
(a)
(b)
Figure 01: (a) Original circuit & (b) Thevenin’s equivalent circuit
Thevenin equivalent voltage is the open circuit (no load) voltage between two output
terminals A& B in a circuit. Thevenin equivalent resistance is the total resistance appearing
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between two output terminals A & B in a given circuit with all sources replaced by their
internal resistances. Thevenin’s equivalent circuit acts same in terms of output voltage and
current as visualized in figure 02.
Figure 02: Output voltage and current (a) original circuit (b) Thevenin’s equivalent circuit
Steps for applying Thevenin’s theorem:
 Step 01: Open the two terminals (remove any load) between which you want to find the
Thevenin equivalent circuit
 Step 02: Determine the voltage (Vth) across the two open terminals
 Step 03: Determine the resistance (Rth) between the two open terminals with all sources
replaced with their internal resistances.
 Step 04: Connect Vth and Rth in series to produce the complete Thevenin equivalent
circuit for the original circuit.
 Step 05: Replace the load removed in Step 01 across the terminals of the Thevenin
equivalent circuit. You can now calculate the load current and load voltage using only
Ohm’s law. They have the same value as the load current and load voltage in original
circuit.
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Procedure:- Realize the circuit on breadboard as shown in figure 01(a)Find out voltage
across and current through terminals A & B in original circuit. Then convert this circuit into
Thevenin equivalent circuit by following the steps of Thevenin’s theorem. Vth and Rth for
original circuit can be found as given in figure 03.
Figure 03: (a) Finding Vth (b) Finding Rth
Observation:- Value of current through and voltage across terminals A & B in both original
and thevenin equivalent circuit is same.
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Answer the following questions:Question 01: When and why Thevenin’s theorem is used?
Question 02: What does Thevenin’s theorem state?
Question 03: What are different steps of Thevenin’s theorem?
Question 04: Draw figure of any complex DC resistive circuit and its Thevenin equivalent
circuit that is observed during this practical in the laboratory.
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Lab E x p e r i m e n t # 11
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Title: Norton’s Theorem.
Objective: Objective of this practical is to learn about Norton’s Theorem; and how this
theorem is used to convert a complex resistive DC circuit into Norton equivalent circuit.
Required Apparatus:- Few resistors, bread board, multimeter, DC voltage sources, DC
current sources, connecting wire etc.
Theory:- Norton’s theorem provides a method for simplifying any two terminal complex
resistive DC circuit to an standard equivalent form. Norton’s equivalent circuit consists of an
equivalent current source and a resistor in parallel. A complex circuit and its Norton
equivalent circuit is shown in figure 01.
(a)
(b)
Figure 01: (a) Original circuit & (b) Norton’s equivalent circuit
Norton equivalent current is the close circuit (no load) current between two output terminals
A & B in a circuit. Norton equivalent resistance is the total resistance appearing between two
output terminals A & B in a given circuit with all sources replaced by their internal
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resistances. Norton’s equivalent circuit acts same in terms of output voltage and current as
visualized in figure 02.
Figure 02: Output voltage and current (a) original circuit (b) Norton’s equivalent circuit
Steps for applying Norton’s theorem:
 Step 01: Short the two terminals (remove any load) between which you want to find the
Norton equivalent circuit
 Step 02: Determine the current (IN) through the two shorted terminals
 Step 03: Determine the resistance (RN) between the two open terminals with all sources
replaced with their internal resistances.
 Step 04: Connect IN and RN in parallel to produce the complete Norton equivalent circuit
for the original circuit.
 Step 05: Replace the load removed in Step 01 across the terminals of the Norton
equivalent circuit. You can now calculate the load current and load voltage using only
Ohm’s law. They have the same value as the load current and load voltage in original
circuit.
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Procedure:- Realize the circuit on breadboard as shown in figure 01(a)Find out voltage
across and current through terminals A & B in original circuit. Then convert this circuit into
Norton equivalent circuit by following the steps of Norton’s theorem. IN and RN for original
circuit can be found as given in figure 03.
(a) RN
Figure 03: (a) Finding RN (b) IN
Observation:- Value of current through and voltage across terminals A & B in both original
and Norton equivalent circuit is same.
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Answer the following questions:Question 01: When and why Norton’s theorem is used?
Question 02: What does Norton’s theorem state?
Question 03: What are different steps of Norton’s theorem?
Question 04: Draw figure of any complex DC resistive circuit and its Norton equivalent
circuit that is observed during this practical in the laboratory.
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Lab E x p e r i m e n t # 12
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Title: Maximum Power Transfer Theorem.
Objective: Objective of this practical is to learn about maximum power transfer theorem; and
verify that how and when maximum power is transferred to a load.
Required Apparatus:- Few resistors, bread board, multimeter, DC voltage source,
connecting wire etc.
Theory:- Maximum power transfer theorem states that “For a given source voltage,
maximum power is transferred from a source to a load when the load resistance is equal to
the internal resistance of source”.
In the figure 01 a voltage source Vs with its internal resistance Rs is connected with load
resistor R L in series. According to maximum power transfer theorem, when Rs=R L,
maximum power will be transferred to R L.
Figure 01: Practical voltage source connected with Load resistor
Circuit current can be given as follows
I
Vs
.....(i)
RS  RL
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Power dissipation at load resistor can be given as follows
PL=I^2 * RL
Procedure:- Realize the circuit on breadboard as shown in figure
02.
Figure 02:
Circuit
Here value of RL is variable. For the verification of maximum power transfer theorem,
value of RL should be set such that (i) RL<Rs (ii) RL=Rs (iii) RL>Rs, then find out
power dissipation at RL in all three cases.
Observation:- For different values of R L, various values of PL are obtained. The maximum
power is dissipated at RL only when RL=Rs as shown in table 01 and figure 03.
RL (ohm)
PL (mWatt)
0
0
Table 01: PL versus
25
RL 50
250
320
75
334
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100
326
125
313
Figure 03: PL versus
RL
Answer the following questions:Question 01: What does maximum power transfer theorem states?
Question 02: What is the condition for maximum power transfer to a load?
Question 03: Draw the circuit diagram that was realized during lab and make a table of
observation for that circuit.
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Lab E x p e r i m e n t # 13
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Title: DC RC Circuit.
Objective: Objective of this practical is to learn about
1. Time Constant (T=RC)
2. Charging and Discharging of Capacitor in RC Circuit
Required Apparatus: - Few resistors, Capacitors, bread board, Multimeter, DC voltage
source, connecting wire etc.
Theory:1. Time Constant: The RC time constant is a fixed time interval that equals the product of
the resistance and capacitance in a series RC circuit. The time constant of series RC
circuit determines the rate at which capacitor charges or discharges. Where time constant,
resistance and capacitance is expressed in seconds, ohms and Farads respectively. It is
symbolized by T(Greek letter tau) and the formula is
Charging and discharging Curves for Capacitor voltage is shown in Figure 01.Charging
curve is shown in Fig1 (a) and Discharging Curve is shown in Fig01 (b)
Figure 01
1. Charging & Discharging in RC Circuit: Capacitor is an element which stores electric
energy by charging the charge on it. It will charge up when it is connected to DC voltage
source. Building up of charges in predictable manner is dependent on resistance and
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capacitance of circuit. Capacitance is a measure of Capacitor’s ability to store charge. Its
formula is given as
The clear picture of Charging of Capacitor is shown in Fig02.As Switch is open so
Initially Capacitor is uncharged and plate A & B both have equal charges shown in
Part (a). When Switch is closed capacitor start charging B plate gain electrons and A
plate losses electrons as a result it becomes more positive with respect to plate B arrow
indicates direction of electrons flow shown in part (b) so part(c) shows capacitor is fully
charged and current is zero while part (d) shows capacitor attains charge even switch is
open.
Figure 02
The clear picture of discharging of capacitor is shown in Fig03.When switch is open
capacitor attains charge it is charged up to 50V shown in part(a).When switch is closed
discharges occurs as indicated by arrows shown in part (b) and capacitor is fully
discharged up to 0V shown in part(c)
Figure 03
Current and voltage values during charging and discharging is depicted from
Fig03.During start of Charging current is maximum and voltage is zero shown Fig03 (a).
and when capacitor is fully charged current becomes zero and voltage becomes maximum
shown in fig03(b) and when capacitor discharges so initially current & voltage is
maximum but gradually they decrease up to 0A current and 0V voltage and in this stage
capacitor is said to be fully discharged shown in fig03(c)
Figure 03
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Procedure: - Realize the circuit on breadboard and on Multism as shown in Figure 03 and
observe current and voltage values during charging and discharging of capacitor.
Observation: - Charging & Discharging of Capacitor is observed & Time Constant is
understand and verified.
Answer the following questions:Question 01: Define Capacitor and its capacitance?
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Question 02: Define RC time Constant? Mention its formula.
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Question 03: Explain Charging and discharging of capacitor in your own words. Explain
reason why current is maximum during start of charging procedure?
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