impedance of electric circuits using the differential equation describing it_first_version
DESCRIPTION
Mathematics behind the generalization of electric circuitsTRANSCRIPT
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Student Number: 223947
1
Abstract—Resistive circuits can easily be decomposed with the
application of Kirchoff’s Current and Voltage laws. Also, the
resistance of such circuits is easy to find out. However, realizing
electric Circuits that include the passive components such as
Capacitors and Inductors by treating the circuits as plain
resistive model is not possible. Capacitors and Inductors bear
unique electrical properties that complicate the scenario. Electric
circuits with the capacitors and inductors can be expressed in
terms of differential equations. Once the circuits are realized in
the set of differential equations, the calculation of rest of the
circuit parameters are easier.
Index Terms—Differential Equations, Impedance, Electric
Circuits
I. INTRODUCTION
HIS paper focuses on the calculation of impedance of an
electric circuit by expressing the elements of the circuits in
terms of differential equation. These set of differential
equations not only help in finding the impedance of the circuits
as well as the transient analysis of the circuits. Hence it is
necessary to understand the electric properties of these
components.
Moving forward, the passive components Capacitor and
Inductor will be explained. The laws that govern the circuit
realization will be put forward and we will see how our passive
components fit in those laws. We then find the circuit
impedance using the differential equation generated.
II. THEORY
A. Passive Components in the circuit
INDUCTOR: An inductor is a passive component which
stores energy in its magnetic fields when the electric current
flows through it. Inductance is measured in terms of Henry
(H). In circuits, its symbol is L. Physically, an inductor is an
electrical conductor wired as a coil. When a time varying
current passes through it, it induces voltage in the coil.
The voltage across an inductor is expressed as the product
of the Inductance value of the inductor and the rate of change
of the current through the inductor. Mathematically,
where L is in Henrys, I in Ampere and t in seconds.
Following is the schematic of an inductor
1 2
CAPACITOR: A capacitor is a charge storing device which
stores charge on two conductors separated by a dielectric. Its
unit is Farad and denoted as C in circuits. Any two electrically
charged conductors that have been separated by a dielectric
holds capacitance in between. Charges are stored across the
plates of the capacitor when a time varying voltage has been to
the capacitor. Mathematically,
where i is in ampere, C in Farads, v in volts and t in seconds.
Following is the schematic of a capacitor
1 2
B. Differential Equation
A differential equation is an equation that relates a function
with its one or more derivatives. Differential Equations can be
of the same variable in a function (Ordinary Differential
Equation) or different variables in a multi dimensional
function (Partial Differential Equation). A differential equation
is signified by its order where order is the highest derivative
that appears in the equation.
C. Basic Circuit Laws
Kirchoff’s Current law states that the sum of current entering
the node in the circuit is equal to the sum of current leaving the
node.
Determination of Impedance of an Electric
Circuit from the Differential Equation
describing the circuit
Prabhat Man Sainju
Department of Electronics
Tampere University of Technology, Tampere, Finland
T
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Student Number: 223947
2
Kirchoff’s Voltage Law states that the sum of voltage drops in
the circuit elements in a closed circuit loop is always equal to
zero.
Ohm’s law gives the voltage drop across an impedance (taking
account of the AC circuits) when current passes through it.
III. REALIZATION OF SERIES RLC CIRCUIT
We now assemble the above circuit elements in a resistive
circuit excited by a source and resolve the circuit in terms of
differential equation.
L1 2
R21
V
C
1
2
Above schematic shows the assembly of Inductor L and
Capacitor C along with the Resistor R in the circuit excited a
voltage source V.
Initially before the circuit is excited by a source, the current
across the inductor and the voltage across the capacitor are
assumed to be zero. This passive circuit shows transient
response as the inductor in the circuit does not allow the
current through it to increase suddenly and the capacitor does
not allow the voltage across to increase suddenly.
We apply Kirchoff’s voltage law in the circuit i.e. finding the
sum of voltage drops across the loop.
Let’s consider an AC source exciting the inductor where the
alternating current is expressed as
And also the voltage source is defined as
We now apply derivative with respect to time to both sides to
eliminate the integral in the equation.
Dividing both sides by L on both side,
Solving the individual derivatives we get,
Substituting these values in equation 1.4 and replacing the
values of current and voltage given by equation 1.2 and 1.3,
we get
Dividing both sides by and taking I common on both sides,
By Ohm’s law we know that the ratio of Voltage over the
Current gives the resistance or the impedance of circuit in this
case. Hence,
Z gives the impedance of the circuit with the resistor, inductor
and capacitor as the passive elements.
IV. REALIZATION OF PARALLEL RLC CIRCUIT
R
2
1
L
1
2
C
1
2I
Above is the schematic for the parallel circuit. We assume the
current source as in equation 1.2 and voltage source as in
equation 1.3. Applying the Kirchoff’s Current Law at the node
2, we get
Expressing above equation in generalized form,
As the earlier procedure, this equation 1.9 can also be
simplified into a differential equation as
Differentiating the voltages and current terms in equation 1.10
as for earlier case yields
It is the complex admittance of the circuit. Impedance of the
circuit is given as
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Student Number: 223947
3
C1
1n
1 2
L
27mH
1 2
V2Vac R
10
2
1
V. EXAMPLES
1. Find the impedance of the electric circuit given below:
Solution:
Impedance of Inductor XL= jωL
= j. 2π.f.L
= j x 6.28 x 60 x 27x103
= j10 Ω
Impedance of Capacitor XC =
=
=
=- j7 Ω
Thus, total impedance of the circuit is given as
= (j10+50-j7) Ω
Z = (50 + j3) Ω
2. Find the impedance of the parallel electric circuit given
below:
XL= jωL
= j10 Ω
XC=1/ jωC
= -j7 Ω
Hence, the admittance of the circuit is given as
= 1/j10 -1/j7 + 1/10
= 1/10 + j3/70
This implies its impedance is
Z =1/Y
= 8.448 + j3.62 Ω
In the above examples, we calculated the impedances for the
series and parallel circuits derived from their differential
equations. AC source has been used as a supply to both of the
circuits. The simplified impedance equation generated from
the differential equation was used in these cases.
VI. CONCLUSION
In this paper, we initially defined what the passive elements
are that are used in the electric circuits. We also presented
forward the mathematical realization of those elements
(capacitor and inductor). It was basically mathematical
modeling of the physical behavior of those elements. We then
formed the ordinary differential equation of second degree out
of those elements when placed in series and parallel
combination. Assuming the sources for both cases to be AC
source, we then simplified the differential equation based on
few trigonometric rules. Finally we came across the solution of
the complex impedance for the series RLC circuit. In case of
the parallel RLC circuit, the calculation resulted in the
complex admittance of the circuit which is actually reciprocal
of the admittance. Thus, we can determine the impedance of
the electric circuit using the differential equation defining it.
Apart from this derivation, impedance can also be derived for
the circuit by performing the Laplace transform.
VII. REFERENCES
[1] James W. Nilsson and Susan A. Riedel, Electric
Circuits. -Wesley Publishing Company, Seventh
Edition.
[2] www.en.wikipedia.org
C
1u1
2L
27mH
1
2
V2Vac R
10
2
1