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

prabhat.sainju@tut.fi

T

Student Number: 223947

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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

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