Table of Contents

**Mcqs of Basic Electronics** **(onilne test)**

**Theory of Basic Electronics**

**What is Electronics** **?**

**Electronics:** The branch of engineering science which deals with current conduction through a vacuum or gas or semiconductor called *electronics. Electronics essentially deals with electronics devices and their utilization. A device is that during which current flows through a vacuum or gas or semiconductor. Such devices have valuable properties that enable them to function and behave because of the friend of man today.

**Importance**. Electronics has gained much importance thanks to its numerous applications in industry. The electronics devices are capable of performing the subsequent functions :

**(i) Rectification. **

The conversion of a.c. into d.c. is named rectification. Electronics devices can convert a.c. power into d.c. power (See Fig. 1.1) with very high efficiency. This d.c. supply is often used for charging storage batteries, field supply of d.c. generators, electroplating, etc

**(ii) Amplification. **

the method of raising the strength of a weak signal is called amplification. Electronics devices can accomplish the work of amplification and thus act as amplifiers (See Fig. 1.2). The amplifiers are utilized in a good sort of way. for instance, an amplifier is employed during a radio set where the weak signal is amplified in order that it are often heard loudly. Similarly, amplifiers are utilized in public address systems, television, etc.

**(iii) Control. **

Electronics devices find wide applications in automatic control. for instance, the speed of a motor, the voltage across a refrigerator, etc. is often automatically controlled with the assistance of such devices.

**(iv) Generation.**

Electronics devices can convert d.c. power into a.c. power of any frequency (See Fig. 1.3). When performing this function, they’re referred to as oscillators. The oscillators are used in a good sort of way. for instance, electronics high-frequency heating is employed for annealing and hardening

**(v) Conversion of sunshine into electricity.**

Electronics devices can convert light into electricity. This conversion of sunshine into electricity is understood as photo-electricity. Photo-electric devices are utilized in Burglar alarms, audio recording on motion pictures, etc.

(vi) Conversion of electricity into light.

(vi) Conversion of electricity into light.

Electronics devices can convert electricity into light. This valuable property is utilized in television and radar

**Atomic Structure**

According to modern world theory, the matter is electrical in nature. All the materials are composed of very tiny particles called atoms. The atoms are the building bricks of all things in the visible world. An atom composed of a central nucleus of positive charge around which tiny negatively charged particles, called electrons revolve in different paths or orbits.

**(1) Nucleus**.

It is the central part of an atom and *composed of protons and neutrons. A proton is a positively charged, while the neutron has the nearly same mass as the proton, but has zero charges. That’s why the nucleus of an atom is positively charged. The sum of protons and neutrons constitutes all weight of an atom and is called atomic weight. It is due to the particles in the extra nucleus (i.e. electrons) have negligible weight as compared to protons or neutrons.

∴ atomic weight = no. of protons + no. of neutrons

**(2) Extra nucleus.**

It is the outside part of an atom and contains only electrons. An electron is a negatively charged particle having nearly zero mass. The charge on an electron is equal but opposite to on a proton. Also, the number of electrons is equal to the number of protons in an atom under normal conditions. That’s why an atom is neutral as a whole. The number of electrons or protons in an atom is known atomic number i.e.

atomic number = no. of protons or electrons in an atom

The electrons in an atom revolve around the nucleus in different orbits or paths. The number and arrangement of electrons in an orbit is found by the following rules :

**(i) **The number of electrons in orbit is given by 2n, 2 where n is the number of the orbit. For example,

- First orbit contains 2 × 12 = 2 electrons
- Second orbit contains 2 × 22 = 8 electrons
- Third orbit contains 2 × 32 = 18 electrons

**(ii)** The last orbit cannot have more than 8 electrons.**(iii)** The last but one orbit cannot have more than 18 electrons

**Structure of Elements**

We know that all atoms are made up of protons, neutrons, and electrons. The difference between many types of elements is because of the different number and arrangement of these particles within their atoms. For example, the structure* of a copper atom is more different from that of a structure of carbon atom and hence the two elements have different properties. The atomic structure can be easily made up if we know the atomic weight as well as the atomic number of the element. Thus taking the case of the copper atom,

Atomic weight = 64

Atomic number = 29

∴ No. of protons = No. of electrons = 29

and No. of neutrons = 64 − 29 = 35

Fig. 1.4 shows the structure of the copper atom. It has 29 electrons which are placed in different orbits as follows. The first orbit always has 2 electrons, the second 8 electrons, the third 18 electrons, and the fourth orbit has 1 electron. The atomic structure of all discovered elements can be shown in this way and the reader is advised to try for a few commonly used elements.

**Energy Of An Electron?**

**Energy Of An Electron:** Since electronics deals with small particles called electrons, these tiny particles require a more detailed study. As discussed before, an electron is a negatively charged particle having nearly zero mass. Some of the important most properties of an electron are :

- Charge on an electron, e = 1.602 × 10−19 coulomb
- Mass of an electron, m = 9.0 × 10−31 kg
- Radius of an electron, r = 1.9 × 10−15 metre

The ratio e/m of an electron is 1.77 × 1011 coulombs/kg. This actually means that the mass of an electron is very small as compared to its charge. It’s because of this property of an electron that it is very mobile and is huge influenced by electric or magnetic fields.

**Energy of an Electron**

An electron rotating around the nucleus possesses two kinds of energies viz. kinetic energy because of its motion and potential energy because of the charge on the nucleus. The net energy of the electron is the sum of these two energies. The energy of an electron increases when its distance from the nucleus increases. Thus, an electron in the second orbit has more energy than the electron in the first orbit; electron in the third orbit has larger energy than in the second orbit. It is clear that electrons in the last orbit must have very high energy as compared to the electrons in the internal orbits. These last orbit electrons play a very important role in finding the physical, chemical, and electrical properties of a material.

**Valence Electrons**

The electrons in the outermost orbit of an atom are called valence electrons. The outermost orbit can carry a maximum of 8 electrons. The valence electrons define the physical and chemical properties of a matter. These valence electrons determine whether or not the material is chemically active; metal or non-metal or, gas or solid. These electrons also explain the electrical properties of a material. On the basis of current conductivity, materials are normally classified into conductors, insulators, and semiconductors. As a rough rule, one can determine the electrical behavior of material from the number of valence electrons as under :

(i) Whenever the number of valence electrons of an atom is less than four (i.e. half of the maximum

eight electrons), the material is normally metal and a conductor. Examples are sodium, magnesium, and aluminum which have one, two, and three valence electrons respectively (See Fig. 1.5).

(ii) Whenever the number of valences shell electrons of an atom is more than four, the material is normally a non-metal and an insulator. Examples are nitrogen, sulfur, and neon which have five, six, and eight valence electrons respectively (See Fig. 1.6).

(iii) Whenever the number of valences shell electrons of an atom is four (i.e. exactly one-half of the maximum 8 electrons), the material has both properties of metal and non-metal and is usually a semiconductor. Examples are carbon, silicon, and germanium (See Fig. 1.7).

**Free Electroncs**

**Free Electrons:** The valence electrons of different materials have different energies. The larger the energy of a valence electron, the weaker it is bound to the nucleus. In some substances, particularly metals, the valence electrons possess huge energy that electrons are very loosely attached to the nucleus. These loosely attached valence electrons motion at random within the material and are known as free electrons. The valence electrons which are very loosely connected or attached to the nucleus are called free electrons.

The free electrons can be easily removed or detached by applying a very little amount of external energy. As a matter of fact, these are the free electrons that determine the electrical conductivity of any matter. On this basis, conductors, insulators, and semiconductors can be explaind as under :

- A
**conductor**is a substance which has a huge number of free electrons. When the voltages are applied across a conductor, the free electrons flow towards the positive terminal of supply, constituting the electric current. - An
**insulator**is a material that has practically no free electrons at room temperatures. Therefore, an insulator can not conduct current under the influence of voltage. - A
**semiconductor**is a material that has very small free electrons at ordinary temperatures. Consequently, under the influence of voltage, a semiconductor normally conducts no current.

**Voltage Source**

Any device that generates voltage output continuously is called voltage source. There are two main types of voltage sources, namely; direct voltage source (DC) and alternating voltage source (AC).

(i) **Direct voltage source**. A device that generates direct voltage output continuously is known as a direct voltage source.

Well, known examples are cells and d.c. generators. An important property of a (DC) direct voltage source is that it keeps the same polarity of the output voltage i.e. positive and negative terminals always remain the same. When load resistance RL is connected across such a source,*current passes from positive to the negative terminal via the load [See Fig. 1.8 (i)]. This is known as direct current because it has just uni-direction. The current has uni-direction as the source remains the same polarity of output voltage. The opposition or resistance to load current inside the d.c. the source is called internal resistance Ri. The equivalent circuit of a direct current source is the generated e.m.f. Eg in series having internal resistance Ri of the source as shown in Fig. 1.8 (ii). Referring to Fig. 1.8 (i), it is clear that:

(ii) **Alternating voltage source**. A device that generates alternating voltage output continuously is known as alternating voltage source e.g. a.c. generator. An important property of the alternating voltage source is that it periodically changes the polarity of the output voltage. When load impedance or resistance ZL is connected across such an (AC) Source, current passes through the circuit that periodically changes in direction. This is called alternating current.

The opposition to load current inside the alternating current. the source is known its internal impedance Zi. The equivalent circuit of an alternating current. the source is the produced e.m.f. Eg (r.m.s.) in series with internal impedance Zi of the source as shown in Fig. 1.9 (ii). Referring to Fig. 1.9 (i), it is clear that :

**Constant Voltage Source**

A voltage source that has very small or low internal *impedance as compared with external load impedance is called as a constant voltage source.

In such a case, the output voltage approximately remains the same when load current changes. Fig. 1.10 (i) illustrates a constant voltage source. It is a direct current source of 6 V with internal resistance Ri = 0.005 Ω. If the load current changes over a wide range of 1 to 10 A, for any of these values, the internal drop across Ri (= 0.005 Ω) is blown than 0.05 volt. so that, the voltage output of the supply is between 5.995 to 5.95 volts. This can be taken as constant voltage compared with the large variations in load current. Fig. 1.10 (ii) shows the graph for a constant voltage source. It may be observed that the output voltage always remains constant in spite of the changes in load current. Thus as the load current changes from 0 to 10 A, the output voltage must remains the same (i.e.V1 = V2 ). A constant voltage source is described as shown in Fig. 1.11.

**Example 1.1.** A lead-acid battery fitted in a truck develops 24V and has an internal resistance of 0.01 Ω. It is used to supply current to headlights or front lights etc. If the total load is equal to 100 watts, find :**(i) **voltage drop in internal resistance**(ii) **terminal voltage

**Solution.**

Generated voltage, Eg = 24 V Internal resistance, Ri = 0.01 Ω Power supplied, P = 100 watts**(i)** Let I will be the load current

**Comments:** It is clear from the example that when the internal resistance of the supply is quite small, the voltage drop in internal resistance is very small. Therefore, the terminal voltage substantially remains fixed and constant, and the source act as a constant voltage source irrespective of load current variations.

**Constant Current Source**

A voltage source that has a very huge internal*impedance as compared with external load impedance or resistance is known as a constant current source. In that case, the load current approximately remains the same when the output voltage changes. Fig. 1.12

(i) illustrates a constant current source. It is a direct current source of 1000 V having internal resistance Ri = 900 kΩ. Here, load RL changes over 3: 1 range from 50 k Ω to 150 k Ω. Over these changes in load RL, the circuit current I is important constant at 1.05 to 0.95 mA or nearly 1 mA. It may be pointed that output voltage V varies nearly in the same 3: 1 range as RL, although load current essentially remains **constant at 1mA. The best example of a constant current source appears in vacuum tube circuits where the tube acts as a generator having internal resistance as large as 1 MΩ. Fig. 1.12 (ii) shows the graph of a constant current source. It is visible that the current remains constant even when the output voltage changes substantially. The following points may be noted regarding the constant current source :

(i) Due to the huge internal resistance of the source, the load current remains essentially constant as the load RL is changed.

(ii) The output voltage changes nearly in the same range as RL, although current remains constant.

(iii) The output voltage V is much smaller than the produced voltage Eg because of the large I Ri drop.

Fig. 1.13 shows the symbol of a constant current source.

Example 1.2. A d.c. source generating 500 V has an internal resistance of 1000 Ω. Find the load current if load resistance is (i) 10 Ω (ii) 50 Ω and (iii) 100 Ω.

Solution.

Generated voltage, Eg = 500 Internal resistance, Ri = 1000 Ω

(i) When RL = 10 Ω

It is clear from the above example that load current is essentially constant since Ri>>RL

.

**Conversion of Voltage Source into Current Source**

Fig. 1.14 shows a constant voltage source with voltage V and internal resistance Ri. Fig. 1.15 shows

its nearly equal current source. It can be easily shown that the two circuits act electrically the same method under all conditions.

**(i) **If in **Fig. 1.14,** the load is open-circuited (i.e. RL → ∞ ), then the voltage across terminal A and B is V. If in Fig. 1.15, the load is open-circuited (i.e. RL → ∞), then all current I (= V/Ri) flows through Ri, yielding voltage across terminals AB = I Ri= V. Note that open-circuited voltage across AB is V for both the two circuits and hence they are electrically equally

(ii) If in **Fig. 1.14,** the load is short-circuited (i.e. RL = 0), the short circuit current is defined by:

If in Fig. 1.15, the load is short-circuited or zero path (i.e. RL= 0), the current I (= V/Ri) bypasses Ri in favor of short-circuiting. It is clearly visible that current (= V/Ri ) is the same for the two circuits and hence both are electrically equivalent.

Thus to convert or transfer a constant voltage source into a constant current source, the following procedure may be applied:

- Put a short-circuit across both two terminals in question (terminals AB in the present case) and determine the short-circuit current. Let it be I. Then I is the current provided by the equivalent current source.
- determine the resistance at the terminals with load removed and sources of e.m.f.s replaced by their internal resistances. Let this resistance be R.
- Then equivalent current source can be shown by a single current source of magnitude I in parallel with resistance R.

Note. To convert a current source of magnitude I in parallel with resistance R into the voltage source,

The voltage of voltage source, V = I R

The resistance of voltage source, R = R

Thus voltage source will be represented as voltage V in series with resistance R.

Example 1.3. Convert the constant voltage source shown in Fig. 1.16 into constant current source.

Solution. The solution involves the following steps :

(i) Put a short across AB in Fig. 1.16 and determine the short-circuit current I. Clearly, I= 10/10 = 1 A

That’s why the equivalent current source has a magnitude of 1 A.

(ii) find the resistance at terminals AB with load *removed and 10 V source changed by its internal resistance. The 10 V source has nearly negligible resistance so that resistance at terminals AB is R = 10 Ω.

(iii) The equivalent current source is a source of 1 A in parallel with a resistance of 10 Ω as shown in Fig. 1.17.

Example 1.4. Convert the constant current source in Fig. 1.18 into equivalent voltage source.

Solution. The solution involves the following steps :

(i) To achieve the voltage of the voltage source, multiply the current of the current source by the internal resistance i.e. Voltage of voltage source = I R = 6 mA × 2 kΩ = 12V

(ii) The internal resistance of the voltage source is 2 k Ω.

The equivalent voltage source is a source of 12 V in series with a resistance of 2 k Ω as shown in **Fig. 1.19.**

**Note.** The voltage source should be represented with the +ve terminal in the direction of the current flow.

**Maximum Power Transfer Theorem **

Whenever the load is connected across a voltage source, power is transferred or moved from the source to the load. The amount of power transferred will depend upon the resistance of the load. If the resistance of load RL is made equal to the internal resistance Ri of any source, then maximum power is transferred or moved to the load RL. This is called as **maximum power transfer theorem** and can be defined as follows: Maximum power is transfer from a source to a load when the load resistance and source resistance are equal. This theorem applies the same to direct current as well as alternating current power.* To prove this theorem numerically, consider a voltage source of producing voltage E and internal resistance Ri and transferring power to a load resistance RL [See Fig. 1.20 (i)]. The current I passes through the circuit is given by :

For this source, produced voltage E and internal resistance Ri are constant. Therefore, the power transferred to the load depends upon RL. In order to determine the value of RL for which the value of P is maximum, it is a must to differentiate eq. (i) w.r.t. RL and set the result equal to 0.

Since (RL+Ri) cannot be zero,

∴ Ri− RL= 0

or RL= Ri

i.e. Load resistance = Internal resistance

Thus, for maximum power delivered, load resistance RL and internal resistance Ri of the source must be equal. Under that situation, the load is said to be matched to the source. Fig. 1.20 (ii) shows a graph of power transferred to RL as a function of RL. It may be noted that the efficiency of maximum power delivery is *50% as one-half of the total produced power is consumed in the internal resistance Ri of the source.

**Maximum Power Transfer Theorem** **Applications.**

Electric power systems never work for maximum power delivered because of very low efficiency and a huge voltage drop between generated voltage and load. However, in the electronics circuits, the maximum power delivered is normally desirable. For instance, in a public address system, it is desirable to have a load (i.e. speaker) “matched” to the amplifier circuit so that there is the maximum deliverance of power from the amplifier circuit to the speaker. In such condition, **efficiency is **sacrificed at the cost of high power transfer.**

Example 1.5. A generator produced 200 V and has an internal resistance of 100 Ω. Find the power transferred to a load of (i) 100 Ω (ii) 300 Ω. Comment on the result.

Solution.

Generated voltage, E = 200 V

Internal resistance, Ri= 100 Ω

(i) When load RL= 100 Ω

Thus, out of 200 W power generated by the generator, only 100W has reached the load i.e.

efficiency is 50% only

Thus, out of 100 watts of power developed by the generator, 75 watts is delivered to the load i.e.

efficiency is 75%.

Comments. Although in the case of RL = Ri a huge power (100 W) is delivered to the load, there is a large wastage of power in the generator. On the other hand, when RL and Ri are not equal.

Electronics devices generate small power. Therefore, if too much efficiency is sought, a huge number of such devices will be connected in series to get the required output. This will distort the output as well as increase the cost and size of the device.

Power transfer is less (75 W) but a smaller part is wasted in the generator i.e. efficiency is very high. Thus, it depends upon a particular condition as to what the load should be. If we want to deliver maximum power (e.g. in amplifiers) irrespective of efficiency, we should make RL = Ri, However, if efficiency is more necessary (e.g. in power systems), then internal resistance of the source should be taken smaller than the load resistance.

**Example 1.6. An audio amplifier generates an alternating output of 12 V before the connection to a load. The amplifier circuit has an equivalent resistance of 15 Ω at the output. What resistance the load required to have to generate maximum power? Also, finf the power output under this condition.**

**Solution**. In order to generate maximum power, the load (e.g. a speaker) should have a resistance of 15 Ω to match the amplifier circuit. The equivalent circuit is shown in Fig. 1.21.

∴ Load required, RL = 15 Ω

**Example 1.7. For the alternating current the generator is shown in ****Fig. 1.22 (i)****, find the value of load so that maximum power is delivered to the load (ii) the value of maximum power.**

(i) In the alternating current system, maximum power is transferred to the load impedance (ZL) when the load impedance is conjugate of the internal impedance (Zi) of the source. Now in the problem, Zi = (100 + j50)Ω.

For maximum power delivered, the load impedance should be conjugate of internal impedance i.e. ZL should be (100 − j50) Ω. This is shown in the dotted line in Fig. 1.22 (ii).

Point out that by making internal impedance and load impedance conjugate, the reactive terms cancel. The circuit then consists of internal and external resistances only. This is quite logical because power is only dissipated in resistances as reactances (XL or XC) dissipated no power

**Thevenin’s Theorem**

Sometimes it is desirable to determine a particular branch current in a circuit as the resistance of that branch is changing while all other resistances and voltages remain constant. For instance, in the circuit shown in Fig. 1.23, it may be required to determine the current through RL for five values of RL, let that R1, R2, R3, and E remain constant. In such condition, the *solution can be obtained readily by using Thevenin’s theorem stated below :

**Thevenin’s theorem state that** **Any two-terminal network having a number of e.m.f. sources and resistances can be changed by an equivalent series circuit using a voltage source E0 in series with a resistance R0 where,**

E0= open-circuited voltage between the two terminals.

R0= the resistance between two terminals of the circuit find by looking “in” at

the terminals with removed load and voltage sources changed by their internal resistances if any.

To understand and learn the usage of this theorem, consider the two-terminal circuit shown in Fig. 1.23. The circuit enclosed in the box can be replaced or changed by one voltage E0 in series with resistance R0 as shown in Fig. 1.24. The response at the terminals AB and A′B′ are the same for the two circuits, independent of the RL values connected across the terminals.

(i) Finding E0. Eo is the voltage which is between terminals A and B of the circuit when load RL is remove. Fig. 1.25 shows the circuit with no load. The voltage drop across R2 is the required voltage E0

.

(ii) Finding R0. R0 is the resistance between terminals A and B with No load and e.m.f. decreased to zero (see Fig. 1.26).

Thus, the value of R0 is found. Once the values of E0 and R0 are found, then the current passes through the resistance of load RL can be determined easily (Refer to Fig. 1.24).

**Procedure for Finding Thevenin Equivalent Circuit**

- (i) Open the two terminals (i.e. remove any load) between which you need to determine the Thevenin equivalent circuit.
- (ii) determine the open-circuit voltage between the two open terminals. It is known as Thevenin voltage E0.
- (iii) Find the resistance between the two open terminals and must all ideal voltage sources shorted and all ideal current sources must be opened (a non-ideal source is replaced by its internal resistance). It is known as Thevenin resistance R0.
- (iv) Connect E0 and R0 in series to develop the Thevenin equivalent circuit between the two terminals under consideration.
- (v) Put the same load resistor removed in step (i) across the terminals of the Thevenin equivalent circuit. The load current can now be determined by using only Ohm’s law and it has the exact same value as the load current in the original circuit.

Example 1.8. Using Thevenin’s theorem, determine the current passes through 100 Ω resistance connected across terminals A and B in the circuit of Fig. 1.27.

Solution.

(i) Finding E0. Eo is the voltage across terminals A and B and removes the 100 Ω resistance as shown in Fig. 1.28.

(ii) Finding R0. R0 is the resistance between terminals A and B with 100 Ω resistance removed and all voltage source are short-circuited as shown in Fig. 1.29.

R0= Resistance at terminals A and B in Fig. 1.29

Therefore, Thevenin’s equivalent circuit will be as shown in Fig. 1.30. Now, the current passes through 100 Ω resistance connected across terminals A and B can be determined by applying Ohm’s law.

Example 1.9. Find the Thevenin’s equivalent circuit for Fig. 1.31.

Solution. The Thevenin’s voltage E0 is the voltage between the terminals of A and B. This voltage is equal to the voltage across R3 because of parallel. It is due to terminals A and B are open-circuited and there is zero current flowing through R2 and hence zero voltage drop across it.

The Thevenin’s resistance R0 is the resistance measured between terminals A and B with load removed (i.e. open at terminals A and B) and voltage source changed by a short circuit.

Therefore, Thevenin’s equivalent circuit will be as shown in Fig. 1.32.

Example 1.10. Calculate the value of load resistance RL to which maximum power may be delivered from the circuit shown in Fig. 1.33 (i). Also determine the maximum power.

Solution. First we determine Thevenin’s equivalent circuit to the left of terminals AB in Fig. 1.33 (i)

The Thevenin’s equivalent circuit to the left of AB terminals in Fig. 1.33 (i) is E0 (= 40 V) in series with R0 (= 73.33 Ω). When RL is connected between A and B terminals, the circuit will as shown in Fig. 1.33 (ii). It is clear that maximum power will be transferred or delivered when

Example 1.11. Calculate the current in the resistor 50 Ω in the network shown in Fig. 1.34.

Solution. We will simplify the given circuit shown in Fig. 1.34 by the multi-use of Thevenin’s theorem. First, we determine Thevenin’s equivalent circuit to the left of *XX.

We can again replace or change the circuit to the left of YY in Fig. 1.35 by its Thevenin’s equivalent circuit. Therefore, the original circuit decreases to that shown in Fig. 1.36.

**What is Nortons Theorem**

Fig. 1.38 (i) shows a circuit enclosed in a box with A and B terminals brought out. The network in the box may have any number of e.m.f. sources and resistors connected in any manner. But according to Norton, the complete circuit behind A and B terminals can be changed by a current source of output IN in parallel with a one resistance RN as shown in Fig. 1.38 (ii). The value of IN is found as mentioned in Norton’s theorem. The resistance RN is the exactly same as Thevenin’s resistance R0. Once Norton’s equivalent is found [See Fig. 1.38 (ii)], then-current passes through any load RL connected across terminals AB can be readily obtained.

Hence Norton’s theorem as applied to direct current circuits may be stated as under Any network having two A and B terminals can be changed by a current source of output IN in parallel with a resistance RN

(i) The output IN of the current source is exactly equal to the current that would passes through AB when A and B terminals are short-circuited.

(ii) RN is the resistance of the network measured between A and B terminals with load (RL) removed and sources of e.m.f. replaced or changed by their internal resistances, if any. Norton’s theorem is converse of Thevenin’s theorem in that Norton equivalent uses a current producer instead of voltage generator and resistance RN (which is the same as R0 ) in parallel with the generator instead of being in series with it. Illustration. Fig. 1.39 illustrates the application of Norton’s theorem. As far as the circuit behind AB terminals is concerne [See Fig. 1.39 (i)], it can be changed by a current source of output IN in parallel with a resistance RN as shown in Fig. 1.39 (iv). The output IN of the current producer or generator is equal to the current that would passes through AB terminals when A and B terminals are short-circuited as shown in Fig. 1.39 (ii). The load R′ on the source when terminals AB are short-circuited is given by :

To determine RN, remove the load RL and change the voltage source by a short circuit because its resistance is assumed 0 [See Fig. 1.39 (iii)].

Thus the IN and RN values are known. The Norton equivalent circuit will be as shown in Fig. 1.39 (iv).

**Procedure for Finding Norton Equivalent Circuit**

- Open the two terminals as the Thevenin theorem (i.e. remove any load) between which we need to find Norton equivalent circuit.
- (ii) make a short-circuit across the terminals under consideration. determine the short-circuit current flowing or passing in the short circuit.
**It is known as Norton’s current IN.** - (iii) Find the resistance between the two open terminals with all ideal current sources opened and all ideal voltage sources shorted (a non-ideal source is replaced by its internal resistance).
**It is called Norton’s resistance RN**. It is easy to see that RN = R0 - (iv) Connect IN and RN in parallel to develop Norton equivalent circuit between the two terminals under consideration.
- make the load resistor removed in step (i) across the terminals of the Norton equivalent circuit. The load current can now be determined by using the current-divider rule. This load current is the same as the load current in the original or first given circuit.

**Example 1.12.** **Using Norton’s theorem, determine the current in 8 Ω resistor in the network shown in** Fig. 1.40 (i).

**Solution.** We shall decrease the network to the left of AB in Fig. 1.40 (i) to Norton’s equivalent. For this reason, we need to find IN and RN.

(i) With removed load (i.e., 8 Ω) and terminals, AB short-circuited [See Fig. 1.40 (ii)], the current that passes through AB is equal to IN. Referring to Fig. 1.40 (ii),

**(ii)** With removed load (i.e., 8 Ω) and battery changed by a short (since its internal resistance is assumed zero), the resistance at AB terminals are equal to RN as shown in Fig. 1.41 (i).

The Norton’s equivalent circuit behind AB terminals is IN (= 3.24 A) in parallel with RN (= 7.4 Ω). When the load (i.e., 8 Ω) is placed across AB terminals, the circuit becomes as shown in Fig. 1.41 (ii). The current source is providing current to two resistors **7.4 Ω and 8 Ω **in parallel.

Example 1.13. **determine the Norton equivalent circuit at terminals X – Y in** Fig. 1.42.

Solution. We shall first determine the Thevenin equivalent circuit and then convert or transfer it to an equivalent current source. This will be then Norton equivalent circuit.

**Finding the Thevenin Equivalent circuit.** To find E0, refer to Fig. 1.43 (i). Since 18 V and 30 V sources are in opposition, the circuit current I is given by :

Fig. 1.44 (iii) shows the Norton equivalent circuit. analyze that the Norton equivalent resistance has the exact same value as the Thevenin equivalent resistance was. Therefore, RN is found exactly the same way.

Example 1.14. **Show that when Thevenin’s equivalent circuit of a network is transferred into Norton’s equivalent circuit, IN= E0/R0 and RN = R0. Here R0 and E0 are Thevenin resistance**

**and Thevenin voltage respectively.**

Solution. Fig. 1.45 (i) shows a network enclosed in a box with two A and B terminals brought out. Thevenin’s equivalent circuit of this network will be as shown in Fig. 1.45 (ii). To determine Norton’s equivalent circuit, we are to determine IN and RN. Referring to Fig. 1.45 (ii)

IN= Current passes through short-circuited AB in Fig. 1.45 (ii)**= E0/R0**

RN= Resistance at terminals AB in Fig. 1.45 (ii)

= R0

Fig. 1.45 (iii) shows Norton’s equivalent circuit. Hence we reached at the following two important conclusions :

(i) To convert or transfer Thevenin’s circuit into Norton’s equivalent circuit,

**Chassis and Ground**

It is the normally practice to mount the electronics components on a metal base** known as chassis.** **For example**, in Fig. 1.46, the resistors and voltage sources are connected to the chassis. As the resistance of chassis is very small or very low, therefore, it gives a conducting path and may be considered as a piece of wire.

It is customary to refer to the chassis as ground. Fig. 1.47 shows the symbol for chassis. It will be seen that all terminals connected to chassis are shown as grounded and represent or shows the same potential. The adoption of this thing (i.e. showing points of the same potential as grounded) often make it easy for the electronics circuits. In our further discussion, we shall most use this scheme.

Mcqs of electronics of the first chapter which are most important and usually this type of Mcqs of electronics comes in job tests not all of them but 2 or 3 from this post. these Mcqs of electronics are prepared for the learners who are now trying to get a job.

- The outermost orbit of an atom may have a maximum of ………….. electrons.

(i)8 (ii) 6 (iii)4 (iv) 3

- When the outermost orbit of an atom has blow than 4 electrons, the matter is generally known as …………..

(i) non-metal (ii) metal (iii) semiconductor (iv) none of above

- The valence or last shell electrons have …………..

(i) very small energy (ii) least energy (iii) maximum energy (iv) none of the above

- A huge number of free electrons exist in …………..

(i) semiconductors (ii) metals (iii) insulators (iv) non-metals

- An ideal voltage supply or source has ………….. internal resistance.

(i) small (ii) large (iii) infinite (iv) zero

- An ideal current source or supply has ………….. internal resistance.

(i) infinite (ii) zero (iii) small (iv) none of the above

- Maximum power is transferred or delivered if the load resistance is equal to ………. of the source.

(i) half the internal resistance (ii) internal resistance (iii) twice the internal resistance (iv) none of the above

- Efficiency at maximum power delivered is …………..

(i) 75% (ii) 25% (iii) 90% (iv) 50%

- When the outermost orbit of an atom has exactly equal to the 4 valence electrons, that matter is generally …………..

(i) a metal (ii) a non-metal (iii) a semiconductor (iv) an insulator

- Thevenin’s theorem replaces or transfers a complicated circuit facing a load by an …………..

(i) ideal voltage source and parallel resistor

(ii) ideal current source and parallel resistor

(iii) ideal current source and series resistor (iv)ideal voltage source and series resistor

- The output voltage supply of an ideal voltage source is …………..

(i) zero (ii) constant

(iii) dependent on load resistance (iv) dependent on the internal resistance

- The current output supply of an ideal current source is …………..

(i) zero (ii) constant

(iii) dependent on load resistance (iv) dependent on internal resistance

- Norton’s theorem replaces or changes a complicated circuit facing a load by an …………..

(i) ideal voltage source and parallel resistor

(ii) ideal current source and parallel resistor

(iii) ideal voltage source and series resistor

(iv) ideal current source and series resistor

- The practical example of the ideal voltage supply is …………..

(i) lead-acid cell (ii) dry cell

(iii) Daniel cell (iv) none of the above

- The electrons speed in a vacuum is ………….. than in a conductor.

(i) less (ii) much more

(iii) much less (iv) none of the above

- Maximum power will be delivered from a source of 10 Ω resistance to a load of …………..

(i) 5 Ω (ii) 20 Ω

(iii) 10 Ω (iv) 40 Ω

- When the outermost orbit of an atom has greater than four electrons, that matter is normally a …………..

(i) metal (ii) non-metal

(iii) semiconductor (iv) none of the above

- Ideal supply having of 5 V in series with 10 k ohm resistance. The current magnitude of the equivalent current source is …………..

(i) 2 mA (ii) 3.5 mA

(iii) 0.5 mA (iv) none of the above

- To achieve Thevenin voltage, you have to…………..

(i) short the load resistor (ii) open the load resistor

(iii) short the voltage source (iv) open the voltage source

- To achieve the Norton current, you have to…………..

(i) short the load resistor (ii) open the load resistor

(iii) short the voltage source (iv) open the voltage source

- The open-circuited voltage at the load RL terminals in a circuit is 30 V. Under the conditions of maximum power delivered, the load voltage will be …………..

(i) 30 V (ii) 10 V (iii) 5 V (iv) 15 V

- Under the conditions of maximum power delivered, a voltage supply is transferring power of 30 W to the load. The power generated by the source is …………..

(i) 45 W (ii) 60 W (iii) 30 W (iv) 90 W

- The maximum power transfer or delivered theorem is used in …………..

(i) electronics circuits

(ii) power system

(iii) home lighting circuits

(iv) none of the above

- The Norton resistance of a circuit is 20 Ω and the shorted-load current is 2 A. If the network is loaded by a resistance may be equal to 20 Ω, the current passes through the load will be…………..

(i) 2 A (ii) 0.5 A

(iii) 4 A (iv) 1 A

- The Norton current is also sometimes known as…………..

(i) shorted-load current

(ii) open-load current

(iii) Thevenin current

(iv) Thevenin voltage

- The outermost orbit of an atom may have a maximum of ………….. electrons.

(i)8 (ii) 6 (iii)4 (iv) 3

- When the outermost orbit of an atom has blow than 4 electrons, the matter is generally known as …………..

(i) non-metal (ii) metal (iii) semiconductor (iv) none of above

- The valence or last shell electrons have …………..

(i) very small energy (ii) least energy (iii) maximum energy (iv) none of the above

- A huge number of free electrons exist in …………..

(i) semiconductors (ii) metals (iii) insulators (iv) non-metals

- An ideal voltage supply or source has ………….. internal resistance.

(i) small (ii) large (iii) infinite (iv) zero

- An ideal current source or supply has ………….. internal resistance.

(i) infinite (ii) zero (iii) small (iv) none of the above

- Maximum power is transferred or delivered if the load resistance is equal to ………. of the source.

(i) half the internal resistance (ii) internal resistance (iii) twice the internal resistance (iv) none of the above

- Efficiency at maximum power delivered is …………..

(i) 75% (ii) 25% (iii) 90% (iv) 50%

- When the outermost orbit of an atom has exactly equal to the 4 valence electrons, that matter is generally …………..

(i) a metal (ii) a non-metal (iii) a semiconductor (iv) an insulator

- Thevenin’s theorem replaces or transfers a complicated circuit facing a load by an …………..

(i) ideal voltage source and parallel resistor

(ii) ideal current source and parallel resistor

(iii) ideal current source and series resistor (iv)ideal voltage source and series resistor

- The output voltage supply of an ideal voltage source is …………..

(i) zero (ii) constant

(iii) dependent on load resistance (iv) dependent on the internal resistance

- The current output supply of an ideal current source is …………..

(i) zero (ii) constant

(iii) dependent on load resistance (iv) dependent on internal resistance

- Norton’s theorem replaces or changes a complicated circuit facing a load by an …………..

(i) ideal voltage source and parallel resistor

(ii) ideal current source and parallel resistor

(iii) ideal voltage source and series resistor

(iv) ideal current source and series resistor

- The practical example of the ideal voltage supply is …………..

(i) lead-acid cell (ii) dry cell

(iii) Daniel cell (iv) none of the above

- The electrons speed in a vacuum is ………….. than in a conductor.

(i) less (ii) much more

(iii) much less (iv) none of the above

- Maximum power will be delivered from a source of 10 Ω resistance to a load of …………..

(i) 5 Ω (ii) 20 Ω

(iii) 10 Ω (iv) 40 Ω

- When the outermost orbit of an atom has greater than four electrons, that matter is normally a …………..

(i) metal (ii) non-metal

(iii) semiconductor (iv) none of the above

- Ideal supply having of 5 V in series with 10 k ohm resistance. The current magnitude of the equivalent current source is …………..

(i) 2 mA (ii) 3.5 mA

(iii) 0.5 mA (iv) none of the above

- To achieve Thevenin voltage, you have to…………..

(i) short the load resistor (ii) open the load resistor

(iii) short the voltage source (iv) open the voltage source

- To achieve the Norton current, you have to…………..

(i) short the load resistor (ii) open the load resistor

(iii) short the voltage source (iv) open the voltage source

- The open-circuited voltage at the load RL terminals in a circuit is 30 V. Under the conditions of maximum power delivered, the load voltage will be …………..

(i) 30 V (ii) 10 V (iii) 5 V (iv) 15 V

- Under the conditions of maximum power delivered, a voltage supply is transferring power of 30 W to the load. The power generated by the source is …………..

(i) 45 W (ii) 60 W (iii) 30 W (iv) 90 W

- The maximum power transfer or delivered theorem is used in …………..

(i) electronic circuits

(ii) power system

(iii) home lighting circuits

(iv) none of the above

- The Norton resistance of a circuit is 20 Ω and the shorted-load current is 2 A. If the network is loaded by a resistance may be equal to 20 Ω, the current passes through the load will be…………..

(i) 2 A (ii) 0.5 A

(iii) 4 A (iv) 1 A

- The Norton current is also sometimes known as…………..

(i) shorted-load current

(ii) open-load current

(iii) Thevenin current

(iv) Thevenin voltage

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