BEEE UNIT WISE IMP QNA, Exercises of Basic Electronics

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1. Explain about passive elements.

(or)

Explain the relationship between active and passive elements.

(or)

Illustrate R, L, C parameters.

Resistance:

When a current flows in a material, the free electrons move through the material and collide with

other atoms. These collisions cause the electrons to lose some of their energy. This loss of energy

per unit charge is the drop in potential across the material. The amount of energy lost by the

electrons is related to the physical property of the material. These collisions restrict the movement

of electrons. The property of a material to restrict the flow of electrons is called resistance, denoted

by R. The symbol for the resistor is shown in Fig. 1.3.

The unit of resistance is ohm (Ω). Ohm is defined as the resistance offered by the material when a

current of one ampere flows between two terminals with one volt applied across it.

According to Ohm’s law, the current is directly proportional to the voltage and inversely

proportignal to the total resistance of the circuit, i.e.

We can write the above equation in terms of charge as follows.

where G is the conductance of a conductor. The units of resistance and conductance are ohm (Ω)

and mho (𝖴) respectively.

When current flows through any resistive material, heat is generated by the collision of electrons

with other atomic particles. The power absorbed by the resistor is converted to heat. The power

absorbed by the resistor is given by

where i is the current in the resistor in amps, and v is the voltage across the resistor in volts. Energy

lost in a resistance in time t is given by

Where v is the volts, R is in ohms, t is in seconds and W is in joules

Inductance:

A wire of certain length, when twisted into a coil becomes a basic inductor. If current is made to

pass through an inductor, an electromagnetic field is formed. A change in the magnitude of the

current changes the electromagnetic field. Increase in current expands the fields, and decrease in

current reduces it. Therefore, a change in current produces change in the electromagnetic field,

which induces a voltage across the coil according to Faraday’s law of electromagnetic induction.

The unit of inductance is henry, denoted by H. By definition, the inductance is one henry when

current through the coil, changing at the rate of one ampere per second, induces one volt across the

coil. The symbol for inductance is shown in Fig. 1.4.

The current-voltage relation is given by ,

where v is the voltage across inductor in volts, and i is the current through inductor in amps. We

can rewrite the above equations as

Integrating both sides, we get

From the above equation we note that the current in an inductor is dependent upon the integral of

the voltage across its terminals and the initial current in the coil, i(0).

The power absorbed by inductor is

The energy stored by the inductor is

2. Describe types of energy sources.

Types of Energy Sources:

There are basically two Types of Energy Sources ; voltage source and current source. These

are classified as

i) Ideal source and

ii) Practical source.

Let us see the difference between ideal and practical sources.

Voltage Source

Ideal voltage source is defined as the energy source which gives constant voltage across its

terminals irrespective of the current drawn through its terminals. The symbol for ideal voltage

source is shown in the Fig. 1.5(a). This is connected to the load as shown in Fig. 1.5(b). At any time

the value of voltage at load terminas remains same. This is indicated by V- I characteristics shown

in the Fig. 1.5 (c).

But practically, every voltage source has small internal resistance shown in series with voltage

source and is represented by R se

as shown in the Fig.1.6.

Because of the R se

voltage across terminals decreases slightly with increase in current and it is

given by expression,

For ideal voltage source, R se

Current Source

Ideal current source is the source which gives constant current at its terminals irrespective of the

voltage appearing across its terminals. The symbol for ideal current source is shown in the Fig.

1.9 (a). This is connected to the load as shown in the Fig. 1.9 (b). At any time, the value of the

current flowing through load IL is same i.e. is irrespective of voltage appearing across its terminals.

This is explained by V-I characteristics shown in the Fig. 1.9 (c).

But practically, every current source has high internal resistance, shown in parallel with current

source and it is represented by R sh

. This is shown in the Fig. 1.10.

Because of R sh,

current through its terminals decreases slightly with increase in voltage at its

terminals.

For ideal current source, R sh

= ∞ and generally not shown.

3. Explain about current division rule and voltage division rule.

Current Division:

In a parallel circuit, the Current Division in all branches. Thus, a parallel circuit acts as a current

divider. The total current entering into the parallel branches is divided into the branches currents

according to the resistance values. The branch having higher resistance allows lesser current, and

the branch with lower resistance allows more current. Let us find the current division in the parallel

circuit shown in Fig. 1.32.

The voltage applied across each resistor is V s

. The current passing through each resistor is given by

where V m

is the voltage across mth resistor, R m

is the resistance across which the voltage is to be

determined and R T

is the total series resistance.

4. Explain about source transformation technique.

Source Transformation Technique:

In solving networks to find solutions one may have to deal with energy sources. basically, energy

sources are either voltage sources or current sources. Sometimes it is necessary to convert a voltage

source to a current source and vice-versa.

5. State and explain Kirchhoff’s laws.

Kirchhoff’s Voltage Law :

Kirchhoff’s voltage law states that the algebraic sum of all branch voltages around any closed path

in a circuit is always zero at all instants of time. When the current passes through a resistor, there is

a loss of energy and, therefore, a voltage drop. In any element, the current always flows from

higher potential to lower potential. Consider the circuit in Fig. 1.11. It is customary to take the

direction of current I as indicated in the figure, i.e. it leaves the positive terminal of the vol- tage

source and enters into the negative terminal.

As the current passes through the circuit, the sum of the voltage drop around the loop is equal to the

total voltage in that loop. Here the polarities are attributed to the resistors to indicate that the

voltages at points a, c and e are more than the voltages at b. d and f respectively. as the current

passes from a to f

Consider the problem of finding out the current supplied by the source V in the circuit shown in

Fig. 1.12. Our first step is to assume the reference current direction and to indicate the polarities

for different elements. (See Fig. 1.13).

By using Ohm’s law, we find the voltage across each resistor as follows.

where V R1,

V

R

and V R

are the voltages across R 1,

R

2

and R 3

, respectively. Finally, by applying

Kirchhoff’s law, we can form the equation

From the above equation the current delivered by the source is given by

Kirchhoff’s Current Law:

Kirchhoff’s Current Law states that the sum of the currents entering into any node is equal to the

sum of the currents leaving that node. The node may be an interconnection of two or more

branches. In any parallel circuit, the node is a junction point of two or more branches.

The above two circuits are equal if their respective resistances from the terminals AB, BC and CA

are equal. Consider the Star Delta Control Circuit in Fig. 3.1(a); the resistance from the terminals

AB, BC and CA respectively are

Similarly, in the delta connected network in Fig. 3.1(b), the resistances seen from the terminalsAB,

BC and CA, respectively, are

Now, if we equate the resistances of Star Delta Control Circuit, we get

Subtracting Eq. 3.2 from Eq. 3.1, and adding Eq. 3.3 to the resultant, we have R I

R

Thus, a delta connection of R 1

, R

2

and R 3

may be replaced by a star connection of R A

, R

B

andR C

as determined from Eqs 3.4, 3.5 and 3.6. Now if we multiply the Eqs 3.4 and 3.5, 3.5 and 3.6,

3.6 and 3.4, and add the three, we get the final equation as under:

In Eq. 3.7 dividing the LHS by R A

, gives R 3

; dividing it by R B

gives R 2

, and doing the same withR C

gives R 1.

From the above results, we can say that a star connected circuit can be transformed into a delta

connected circuit and vice-versa.

From Fig. 3.2 and the above results, we can conclude that any resistance of the delta circuit is equal

to the sum of the products of all possible pairs of star resistances divided by the opposite resistance

8. State and explain superposition theorem.

Superposition Theorem:

This theorem states that in any linear network containing two or more sources, the response in any

element is equal to the algebraic sum of the responses caused by individual sources acting alone,

while the other sources are non-operative; that is, while considering the effect of individual

sources, other ideal voltage sources and ideal current source in the network are replaced by short

circuit and open circuit across their terminals. This theorem is valid only for linear systems. This

theorem can be better understood with a numerical example.

9. State and explain thevenin’s throrem.

Thevenin’s theorem:

This theorem states that any two terminal linear network having a number of voltage current sources

and resistances can be replaced by a simple equivalent circuit consisting of a single voltagesource in

series with a resistance, where the value of the voltage source is equal to the open circuit voltage

across the two terminals of the network, and resistance is equal to the equivalent resistancemeasured

between the terminals with all the energy sources are replaced by their internal resistances.

10.State and explain maximum power transfer theorem.

12.State and explain reciprocity theorem.

Reciprocity theorem:

In any linear bilateral network containing the response at any branch or transformation ratio is

same even after interchanging the sources. i.e. V/I 1 =V/I 2

13.Explain the behavior of AC through pure resistance.

Consider an AC circuit with a pure resistance R as shown in the figure. The alternating voltage v

is given by

𝑚

sin(𝜔𝑡)

The current flowing in the circuit is i. The voltage across the resistor is given as V R

which is

the same as v.

Using ohms law, we can write the following relations

i = =

𝑚

sin(𝜔𝑡)

i = i𝑚sin(𝜔𝑡)

Where

i 𝑚

𝑚

in a pure resistive circuit, the voltage and current arein phase. Hence the voltage and current

waveforms and phasors can be drawn as below.

Inductive reactance

The inductive reactance X L

is given as

X𝐿 = 2 𝜋ƒ𝐿

i𝑚

𝑣 𝑚

𝑋 𝐿

15.Explain the behavior of AC through pure capacitance.

AC circuit with a pure capacitance

Consider an AC circuit with a pure capacitance C as shown in the figure. The alternating voltage

v is given by

𝑣 = 𝑣𝑚 sin

0

The current flowing in the circuit is i. The voltage across the capacitor is given as V C

which is

the same as 𝑣.

We can find the current through the capacitor as follows

q = 𝐶𝑣𝑚 sin

𝑑𝑞

= CV

m

𝜔cos(𝜔𝑡)

𝑑𝑡

i = 𝜔𝐶𝑣 𝑚

cos(𝜔𝑡)

i = 𝜔𝐶𝑣𝑚sin(𝜔𝑡 +

i = i 𝑚

sin(𝜔𝑡 +

i 𝑚

𝑚

=> X

𝐶

𝑚

In a pure capacitive circuit, the

current leads the voltage by 90

0

.

𝜔𝐶

i

16.Explain the behavior of AC through R-L series circuit.

R-L Series Circuit:

Consider an AC circuit with a resistance R and an inductance L connected in series as shown

in the figure. The alternating voltage v is given by

𝑚

s in( )

The current flowing in the circuit is i. The voltage across the resistor is V R

and that across the

inductor is V L

V

R

=IR is in phase with I

V

L

=IX

L

leads current by 90 degrees

With the above information, the phasor diagram can be drawn as shown.

The current I is taken as the reference phasor. The voltage VR is in phase with I and the

voltage V L

leads the current by 90⁰. The resultant voltage V can be drawn as shown in the

figure. From the phasor diagram we observe that the voltage leads the current by an angle Φ or

in other words the current lags behind the voltage by an angle Φ.

The waveform and equations for an RL series circuit can be drawn as below.

𝑚

s i n(𝜔𝑡 )

𝐼 = 𝐼𝑚s i n(𝜔𝑡 − ∅)

From the phasor diagram, the expressions for the resultant voltage V and the angle Φ can be

derived as follows.