Part:1
Construct the circuit shown in figure1 on Multisim with R = 450 ohm.
Voltage -Current Relationship in a DC Circuit:
Table:1
For R= 470 ohm
Voltage
1v
2v
3v
4v
5v
6v
7v
8v
Current
2.128mA
4.255mA
6.383mA
8.511mA
10.63mA
12.76mA
14.89mA
17.02mA
1
Table:2
For R= 2.2k ohm
Voltage
1v
2v
3v
Current
0.4545mA 0.9090mA 1.364mA
4v
5v
6v
7v
8v
1.818mA
2.273mA
2.727mA
3.182mA
3.636mA
Table:3
For R= 10k ohm
Voltage
1v
2v
3v
4v
5v
6v
7v
8v
Current
100uA
200uA
300uA
400uA
500uA
600uA
700uA
800uA
2
Conclusion:
Q.1 What is the nature of the graphs plotted?
The nature of the graphs plotted is a voltage-current relationship graph. Each point on the graph represents
a specific current value measured for a constant voltage. The graph is linear as it obey Ohm's law, meaning
that the current is directly proportional to the voltage.
Q.2 What can you determine from the graphs?
A voltage-current relationship graph, also known as an IV curve, can provide valuable information about
the behavior and properties of an electrical component or system. Here are some things that can be
determined from an IV curve:
Resistance, conductance, non-linearity, Operating rang, Breakdown Voltage , Power dissipation.
Q.3 What do the graphs imply with regards to Ohm's law?
The graphs imply that the component in the circuit obeys Ohm's law, which states that the current flowing
through a resistor is directly proportional to the voltage applied across it. The fact that the voltage-current
relationship is a straight line passing through the origin confirms that the component behaves in a linear
and predictable manner, making it useful for many practical applications.
Part:2
Construct the RC circuit in Multisim.
3
Table:1
Time (s)
Charge (Vc)
Discharge (Vc)
0
0v
0v
25
3.936 v
6.075 v
50
6.322 v
3.688 v
75
7.769 v
2.331 v
100
8.647 v
1.453 v
125
9.181 v
0.890 v
150
9.504 v
0.547 v
175
9.700 v
0.351 v
200
9.818 v
0.283 v
225
9.890 v
0.121 v
250
9.934 v
0.066 v
275
9.961 v
0.030 v
300
9.977 v
0.022 v
Time (s)
Charge (Ic)
Discharge (Ic)
0
0A
20 mA
25
7.878 uA
12.120A
50
12.651 uA
7.367 mA
75
15.536uA
4.472 mA
100
17.291 uA
2.716 mA
125
18.363 uA
1.651 mA
150
19.005 uA
0.9957 mA
175
19.398 uA
0.6039 mA
200
19.787 uA
0.3663 mA
225
19.855 uA
0.2221 mA
250
19.917 uA
0.1347 mA
275
19.954uA
0.0817 mA
300
19.93uA
0.0495 mA
Table:2
4
Vc Calculation While Charging the capacitor :
Given:
Resistor (R) = 500kΩ
Capacitor (C) = 100μF
DC voltage (V) = 10V
Using the formula for the voltage across the capacitor while charging:
Vc = Vs * (1 - exp(-t/(R*C)))
At t = 25 seconds:
Vc = 10 * (1 - exp(-25/(500000*0.0001))) = 3.934V
At t = 50 seconds:
Vc = 10 * (1 - exp(-50/(500000*0.0001))) = 6.321V
At t = 75 seconds:
Vc = 10 * (1 - exp(-75/(500000*0.0001))) = 7.768V
Vc Calculation while Discharging the capacitor:
Given:
Resistor (R) = 500kΩ
Capacitor (C) = 100μF
DC voltage (V) = 10V
Using the formula for the voltage across the capacitor while discharging:
Vc(t) = V * exp(-t/(R*C))
5
At t = 25 seconds:
Vc = 10 * exp(-25/(500000*0.0001)) = 6.065V
At t = 50 seconds:
Vc = 10 * exp(-50/(500000*0.0001)) = 3.678V
At t = 75 seconds:
Vc = 10 * exp(-75/(500000*0.0001)) = 2.231V
Ic Calculation while Charging the capacitor.
At t = 25s:
Vc(25) = 10V * (1 - e^(-25 / (500000 * 0.0001))) = 4.96V
Ic(25) = Vc(25) / R = 4.96V / 500000Ω = 0.000007869A = 7.869μA
At t = 50s:
Vc(50) = 10V * (1 - e^(-50 / (500000 * 0.0001))) = 9.27V
Ic(50) = Vc(50) / R = 9.27V / 500000Ω = 0.000012642A = 12.642μA
At t = 75s:
Vc(75) = 10V * (1 - e^(-75 / (500000 * 0.0001))) = 9.99V
Ic(75) = Vc(75) / R = 9.99V / 500000Ω = 0.000015537A = 15.537μA
Ic Calculation while Discharging the capacitor:
At t = 25s:
Vc(25) = 9.99V * e^(-25 / (500000 * 0.0001)) = 4.93V
Ic(25) = Vc(25) / R = 4.93V / 500000Ω = 0.00001226A = 12.26mA
At t = 50s:
Vc(50) = 9.99V * e^(-50 / (500000 * 0.0001)) = 2.42V
Ic(50) = Vc(50) / R = 2.42V / 500000Ω = 0.00000484A = 7.357mA
6
At t = 75s:
Vc(75) = 9.99V * e^(-75 / (500000 * 0.0001)) = 1.18V
Ic(75) = Vc(75) / R = 1.18V / 500000Ω = 0.00000236A = 4.462mA
Calculate the Time Constant of Circuit :
R=500k and C= 100u
Solution:
τ = R x C = (500 x 10^3 Ω) x (100 x 10^-6 F)
τ=
50 seconds
7
Conclusion:
QNo:1 Compare theoretical and measured values of voltage and currents for the capacitor
charging and discharging.
Voltages on Capacitor :
Calculated
Measured
Time (s)
Charging (Vc)
Discharging (Vc)
Charging (Vc)
Discharging (Vc)
25
3.934 v
6.065 v
3.936 v
6.075 v
50
6.321 v
3.678 v
6.322 v
3.688 v
75
7.768 v
2.231 v
7.769 v
2.331 v
Current on Capacitor:
Calculated
Time (s)
25
50
75
Charging (Ic)
7.869 uA
12.642 uA
15.537uA
Discharging (Ic)
12.130mA
7.357 mA
4.462 mA
Measured
Charging (Ic)
7.869 uA
7.869 uA
12.642 uA
Discharging (Ic)
12.120mA
7.367 mA
4.472 mA
Q.2 Discuss the charging and Discharging of capacitor.
Charging of Capacitor:
When a capacitor is connected to a voltage source through a resistor, a current flows through the circuit,
and the capacitor begins to charge up. Initially, the capacitor has no charge, and its voltage is zero. As the
capacitor charges up, the voltage across its plates increases, and the current through the circuit decreases.
The rate of charging is determined by the time constant of the circuit, which is equal to the product of the
resistance and the capacitance. The time constant represents the time it takes for the capacitor to charge up
to 63.2% of the source voltage.
8
Discharging of Capacitor:
When a charged capacitor is disconnected from a voltage source and connected to a resistor, it
discharges through the resistor. Initially, the capacitor has a voltage equal to its charged voltage,
and a current flows through the resistor and the capacitor. As time passes, the voltage across the
capacitor decreases, and the current through the resistor and capacitor decreases accordingly. The
rate of discharging is also determined by the time constant of the circuit, which is equal to the
product of the resistance and the capacitance. The time constant represents the time it takes for the
capacitor to discharge to 36.8% of its charged voltage.
Q.3 Comment on reasons any errors encountered.
No any error encountered.
Q.4 What is the circuit time constant? What does this mean? Approximately how long does
it take the voltage and current to reach their final values?
The time constant of a circuit is a measure of the rate at which a capacitor charges or discharges
in the circuit. It is defined as the product of the resistance (R) and the capacitance (C) of the
circuit, and is usually denoted by the symbol tau (τ).
The time constant (τ) represents the time it takes for the voltage or current in the circuit to reach
approximately 63.2% of its final value during charging or discharging. It is also equal to the time
it takes for the capacitor to charge or discharge to approximately 63.2% of its final voltage. The
time constant is an important parameter in circuit design because it determines how quickly a
circuit responds to changes in input signals or how quickly a capacitor charges or discharges in
response to changes in the circuit.
The time it takes for the voltage or current to reach its final value depends on the value of the time
constant and the initial conditions of the circuit. In general, it takes about five time constants for
the voltage or current to reach approximately 99% of its final value. For example, if the time
constant of a circuit is 1 second, it would take approximately 5 seconds for the voltage or current
to reach its final value. However, this is an approximation and the actual time it takes for the
voltage or current to reach its final value may vary depending on the circuit components and the
specific conditions of the circuit.
9
Capacitor voltage graph while charging:
Capacitor voltage graph while Discharging:
10
Capacitor Current graph while charging:
Capacitor Current graph while Discharging:
11
Part:3
Procedure
1.Construct the RL circuit in Multisim:
3. Measure the current and voltage when the input voltage is high (Switch closed) and low
(Switch open):
High (Switch closed)
low (Switch open)
Current IL
Voltage VL
Current IL
Voltage VL
440 mA
8.06 pv
130.917 nA ≈0A
24v
12
2. Use a square wave to simulate the switch being opened and closed (note you will have to
calculate the circuit time constant to determine a suitable period of square wave).
Time constant of circuit:
Req=54.54 ohm and L= 500mh:
τ = Req /L = (54.54) / (500*10-3)
τ = 9.16 ms
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4. Plot the current and voltage wave form on same axis:
5. Calculate the theoretical values for current and voltage when the input voltage is high (switch
closed) and low (switch open) using the equations provided in Figure 4.
Time constant of circuit:
Req=54.54 ohm and L= 500mh:
τ = Req /L = (54.54) / (500*10-3)
τ = 9.16 ms
Current IL
High (switch closed)
I=
I=
𝑉𝑏
[1 −
𝑅
24
54.54
Low (switch open)
𝑒 (−𝑡/(𝐿/𝑅)) ]
I=
{ L/R = τ
[1 − 𝑒 (−1/(9.16∗10
I= 440.04 mA
−3 ))
]
I=
𝑉𝑏
[
𝑅
𝑒 (−𝑡/(𝐿/𝑅)) ]
24
54.54
{ L/R = τ
[1 − 𝑒 (−0/(9.16∗10
−3 ))
]
I= 0 A
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Voltage VL
High (switch closed)
Low (switch open)
V= Vb [𝑒 (−𝑡/(𝐿/𝑅)) ]
V= Vb[ 𝑒 (−𝑡/(𝐿/𝑅)) ]
{ L/R = τ
V = 24 [𝑒 (−1/(9.16∗10
−3 ))
{ L/R = τ
V= 24 [ 𝑒 (−0/(9.16∗10
]
V= 2.785*10-47v
−3 ))
]
V= 24v
QNo:1 Compare theoretical and measured values of voltage and currents for the inductor
charging and discharging.
Voltage on Inductor:
Calculated
Measured
Charging Vc
Discharging Vc
Charging Vc
Discharging Vc
2.785*10-47 v
24v
8.06 pv
24v
Current on Inductor:
Calculated
Measured
Charging Ic
Discharging Ic
Charging Ic
Discharging Ic
440.04 mA
0A
440 mA
130.917 nA ≈0A
QNo:2 Discuss the charging and discharging characteristics of the inductor.
Charging Characteristics:
When an inductor is connected to a DC voltage source, it initially acts as a short circuit, and the current
through the inductor increases rapidly. However, as the current increases, the magnetic field around the
inductor also increases, and the inductor starts to oppose the current flow. This opposition to current flow
is known as inductive reactance.
15
The rate at which the inductor stores energy is proportional to the rate of change of current. Therefore, the
inductor's charging time depends on the inductance and the resistance of the circuit. Once the inductor is
fully charged, the current through the inductor becomes steady, and the magnetic field around the inductor
stabilizes.
Discharging Characteristics:
When the voltage source is removed from the inductor, the magnetic field collapses, and the energy stored
in the magnetic field is released. This energy is converted back into electrical energy, and the inductor
generates a voltage that opposes the current flow. This opposition to current flow is known as self-induction.
The rate at which the energy is released depends on the inductance and the resistance of the circuit. The
discharging time of the inductor is also proportional to the time constant of the circuit, which is equal to the
inductance divided by the resistance.
In summary, the charging and discharging characteristics of an inductor depend on its electrical properties
and the circuit in which it is connected. The inductor acts as a short circuit during charging and generates a
voltage that opposes the current flow during discharging. The charging and discharging times are
determined by the inductance, resistance, and time constant of the circuit.
QNo:3 Comment on reasons for any errors encountered.
No any error encountered.
QNo:4 What is the circuit time constant and what does this mean?
The circuit time constant is a parameter that describes the behavior of a circuit in response to a sudden
change in voltage or current. It is defined as the product of the resistance and the capacitance or inductance
of the circuit, depending on the type of circuit.
Note: Time constant for lab experiment is 9.167 ms.
The circuit time constant is an important parameter that is used to analyze and design circuits. It determines
the rate at which a circuit responds to changes in voltage or current and can be used to predict the behavior
of the circuit over time. The time constant is also useful for determining the bandwidth and frequency
response of the circuit.
QNo:5 What is the steady state current for the circuit?
Steady-state current is the current that flows through a circuit after all the transient effects have decayed
and a stable operating condition has been reached. In other words, it is the current that flows through a
circuit when all the voltages and currents in the circuit have stabilized and are no longer changing. The
steady-state current is a fundamental property of a circuit and is often used to analyze and design electronic
16
circuits. It is typically the current that is of interest in determining the behavior of a circuit under normal
operating conditions.
In order to determine the steady-state current for a circuit, we can use the principles of Ohm's law and
Kirchhoff's laws. Ohm's law states that the current through a resistor is proportional to the voltage across
it, while Kirchhoff's laws state that the sum of the currents entering a node must be equal to the sum of the
currents leaving the node, and that the sum of the voltages around a closed loop must be zero.
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