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The correct choice is (c) HUMAN ear RESPONSE is logarithmic

The best explanation: The human HEARING scale is logarithmic in nature. For doubling PERCEIVED intensity of sound, the sound POWER must be increased by 10 times. That means the gain of amplifier which controls sound intensity must have gain of 10 for doubling perceived intensity of sound which is in a bell and in 10 decibel scale.

The correct answer is (c) 70.69%

Explanation: USING CLASS B efficiency:

FORMULA:

%n = 78.54 (Vpout / VCC).

Given

Vcc = 20Vp out.

VL = 18V,

RL = 16V,

%= 78.54( 18V/20V),

Efficiency= 70.69%.

Right CHOICE is (c) cavities

Best explanation: Cavities (inverted FINS) embedded in a heat source are the regions formed between adjacent fins that STAND for the essential PROMOTERS of nucleate boiling or condensation. These cavities are usually utilized to extract heat from a variety of heat generating bodies to a heat sink.

Correct choice is (a) Increase the BASE thickness.

The BEST I can explain: Spreading resistance occurs when thermal energy is transferred from a minute area to a larger area in a SUBSTANCE with finite thermal conductivity. In a heat sink, this MEANS that heat does not distribute uniformly through the heat sink base. The spreading resistance occurrence is shown by how the heat travels from the heat source location and causes a large temperature gradient between the heat source and the edges of the heat sink. This means that some fins are at a LOWER temperature than if the heat source were uniform from corner to corner of the base of the heat sink.

Right answer is (b) One transistor

To explain: In Class A amplifier, if the collector current flows all TIMES during the full CYCLE of the input signal, the power amplifier is KNOWN as class A power amplifier. It is less used for higher power output stages, as it has poor efficiency. So it consists of only one transistor.

Right ANSWER is (c) Less Bandwidth

Explanation: GAIN bandwidth product remains constant for any device. So a device with larger gain will have a NARROW bandwidth and vice VERSA. Gain decreases for the high frequency is because of the same phenomenon.

Correct choice is (c) Voltage controlled voltage source

The BEST I can explain: The term op-amp refers to a voltage –controlled voltage source. The input IMPEDANCE is very high (approximately infinity) and the output impedance is very LOW. An op-amp without feedback measures the input voltage and puts out an output voltage proportional to the input. The ideal operational amplifier voltage is maintained constant. It is controlled by input voltage.

Right option is (a) True

The best I can explain: Copper has AROUND twice the thermal CONDUCTIVITY of aluminum, around 400 W/M K for pure copper. Its MAIN applications are in industrial facilities, power plants, SOLAR thermal water systems, HVAC systems, gas water HEATERS, forced air heating and cooling systems, geothermal heating and cooling, and electronic systems.

Right answer is (a) IC increases

The best I can explain: IBis directly PROPORTIONAL to Ic. The IC CURVE is linearly distributed when using IB as a parameter, and exponentially distributed using VBE as a parameter. So when VBE is CONSTANT, the transistor current IC is almost linear with respect to IB.

The CORRECT ANSWER is (a) Gives distorted output

The explanation: The disadvantage of impedance matching is that it gives distorted output. The Power amplifiers generally USE TRANSFORMER COUPLING because the transformer permits impedance matching.

The correct CHOICE is (b) At the middle

The best explanation: By referring to the output CHARACTERISTICS of the class A amplifier operation, the Q-point is placed exactly at the centre of the AC load LINE and the transistor conducts for every point in the INPUT waveform. Therefore, it lies in the middle of d.c. load line.

Right option is (a) Transformer

Explanation: In Transformer coupling, audio TRANSFORMERS are used. This REQUIRES a larger core and more turns, making a more expensive transformer coupling. Transformer coupling is very WIDELY used at radio FREQUENCIES, but it is also practical at audio frequencies, SAY 30 to 30,000 Hz.

Correct option is (b) To separate bias of one stage from another

The best explanation: In RC or transformer COUPLING, a capacitor / transformer is used as coupling device which connects output of first stage with INPUT of second stage. Its FUNCTION is to pass the a.c signal and blocks d.c. bias VOLTAGE.

The correct OPTION is (d) 0.12V

The best I can EXPLAIN: Given: A =100;β = 0.03

Vi = 40mV

To determine:feedback voltage {Vf} for NEGATIVE feedback amplifier;

Formula Used:Vf =β * Vo

Vf =0.03*(100*40*10-3)

Vf =0.12V

Therefore, the feedback voltage is 0.12V for given parameters of negative feedback amplifier.

Right answer is (d) 8.33%

The explanation: 3 VOLT is INPUT

0.25 is distortion in input for 3 volt

then (0.25/3)*100 = 8.33%.

The correct choice is (a) is INCREASED

For explanation: The gate source junctions are reverse biased as a RESULT depletion regions from which extend to the bar by changing gate to source voltage; effective cross sectional area decreases with the function of the gate to source voltage. With INCREASING the cross section area of the channel, the drain current is increased.

The correct answer is (b) pentode

The EXPLANATION: The pentode consists of an evacuated glass envelope containing five electrodes in this order: a cathode heated by a filament, a control grid, a screen grid, a suppressor grid, and a plate (anode).

Similar to a pentode valve, with which the JUNCTION FET greatly RESEMBLES in its operational characteristics, the simplest way of calculating STAGE gain is by the RELATIONSHIP –

A = gfs . RL

Right answer is (c) 60dB

The explanation is: The total gain of a differential AMPLIFIER with ACTIVE load is the sum of the multiple stages when the gain is CALCULATED is decibels. Given, AV1 = 10dB, AV2 = 20dB and AV3 = 30dB

AV = AV1 + AV2 + AV3 = 10dB + 20dB + 30dB = 60dB.

Right choice is (c) Electrical size of the coupling capacitor

The EXPLANATION: Xc = the electrical size of coupling capacitor

Relation of Xc with the signal frequency (f):

Xc = 1 \Xcis inversely proportional to f.

At low frequencies, Xc becomes very large; output reactance of capacitor increases.

The VOLTAGE across load resistance also reduces because some voltage drop TAKES place across Xc.

Thus output voltage reduces. THEREFORE GAIN is very low.

Correct ANSWER is (b) CONDUCTIVE and convection heat transfer

The explanation: The main purpose of a heat sink is to expel heat from a generating source. Heat sinks work through the process of conductive and convection heat transfer. Heat sinks are a passive form of COOLING, as they have no moving parts and REQUIRE no power. In most cases, heat sinks are USED in conjunction with fans.

Correct option is (b) INCREASE

Best explanation: An AC signal amplifier circuit is mainly used to stabilize the DC biased INPUT voltage to the amplifier and thus only amplify the REQUIRED AC signal. This stabilization is achieved by the use of an Emitter Resistance which provides the required amount of automatic biasing needed for a COMMON emitter amplifier. If the emitter resistor is open, then the input ac voltage will eventually increase.

The CORRECT option is (a) to provide cooling to the processor

To explain I WOULD say: The heat sink has a thermal conductor that CARRIES heat away from the CPU into fins that provide a large SURFACE area for the heat to dissipate throughout the rest of the computer, thus cooling both the heat sink and processor.

The correct choice is (B) LEVEL of non- linear distortion decreases

The explanation: Non- linear distortion generally happens because of ESSENCE of non linearity in the exchange qualities of transistor.

If a large amplitude signal is connected to an amplifier to extend the operation of a device BEYOND the linear operational range, then it also generates the non – linear distortion.

The effective input to an amplifier starts to reduce as the negative feedback is introduced; which eventually helps it to OPERATE in the linear region.

This ability of negative feedback turns out to be beneficial in minimizing the level of non-linear distortion.

Correct answer is (c) 5.76

To elaborate: The total gain of a differential amplifier with active load can be CALCULATED by the formula given below. Where, AV = overall gain, AV1 = VOLTAGE gain of FIRST STAGE and AV2 = voltage gain of SECOND stage.

AV = AV1 × AV2 = 4.8 × 1.2 = 5.76

The correct answer is (b) Loading EFFECT of the next stage

The best explanation: The output of first amplifier stage is the input to next stage. In this way OVERALL VOLTAGE gain can be increased, when NUMBER of amplifier stage is used in succession, it is called a multistage amplifier. The load of first amplifier is the input resistance of the second amplifier. Thus overall gain is reduced.

The correct option is (a) db

Best explanation: Db SCALE is LOGARITHMIC. Voltage gain increases EXPONENTIALLY with frequency so using a linear scale means that we need to work with large VALUES of gain, corresponding to SMALL values of frequency.

The correct answer is (d) 31db

To elaborate: The overall gain of a multistage AMPLIFIER is given as the PRODUCT of the gain of the individual stages.

Gain (A) = A1 * A2 * A3 ……* An

Alternately, if the gain of each STAGE is given in db

The overall gain of the amplifier is the sum of gain of each stage

Gain in db = A1 + A2 + A3 (db)

= 10+6+15 = 31db.

Right option is (b) Economy

For explanation: It uses the resistor and capacitor which are not expensive so the cost is low. But it has poor impedance MATCHING because its output impedance is several times larger than the DEVICE; at its TERMINAL end. It is unsuitable for low frequency application.

The correct answer is (d) DECREASING the fin aspect ratio & Using more conductive material

The EXPLANATION is: Fin efficiency is one of the parameters which formulate a higher thermal conductivity material significant. A fin of a HEAT sink may be measured to be a flat PLATE with heat flow in one END and being dissipated into the surrounding fluid as it travels to the other.As heat flows through the fin, the grouping of the thermal resistance of the heat sink impeding the flow and the heat lost due to convection, the temperature of the fin and, consequently, the heat transfer to the fluid, will decrease from the bottom to the end of the fin. Fin efficiency is defined as the actual heat transferred by the fin, separated by the heat transfer were the fin to be isothermal.

Right answer is (a) 50%

Easiest explanation: USING the transformer coupling technique, the efficiency of an amplifier can be enhanced to a great extent. The coupling transformer provides good impedance MATCHING between the LOAD and output, and it is the main REASON behind the improved efficiency. Therefore, Its efficiency is 50%.

Correct option is (c) voltage gain will decrease

Easiest explanation: In common SOURCE JFET, The MAIN purpose of a source bypass CAPACITOR is to provide additional gain at AC. Here, the common-source JFET has a voltage gain of 10; which INCLUDES the additionalAC gain provided. If the source bypass capacitor is removed, then the voltage gain will eventually decrease.

Right answer is (c) Class C

Easiest explanation: The transistor biasing gives a MUCH IMPROVED efficiency of around 80% to the class C amplifier, it introduces a very HEAVY distortion of the output signal. Therefore, class C amplifiers are not suitable for use as AUDIO amplifiers.

The CORRECT choice is (d) CLASS D

To explain I would SAY: The class D amplifier is nonlinear switching amplifiers or PULSE WIDTH modulation amplifiers. These amplifiers are called as digital amplifiers.

The CORRECT answer is (C) INCREASES gain – BANDWIDTH product

To explain: The gain-bandwidth product in a negative amplifier remains constant. Reducing the closed loop gain (>=1) increases the feedback factor and increases the bandwidth.Thus the gain bandwidth product remains constant.

The correct option is (a) Increase VOLTAGE gain

For explanation I would say: A circuit having a SINGLE transistor configuration does not provide SUITABLE BANDWIDTH or gain. The purpose of a differential amplifier with ACTIVE load is to provide an increase in the voltage gain. The total gain of a differential amplifier with active load is the product of the voltage gains of the discrete stages.

Right choice is (b) saturation, cut off

For explanation: Q-point is generally TAKEN to be the intersection point of LOAD LINE with the output CHARACTERISTICS of the transistor. That’s why; the linear operating region of a transistor must LIE along the load line below saturation and above the cut off.

Correct ANSWER is (c) DECIBELS

To elaborate: The voltage GAIN is measured in terms of decibels. The TOTAL voltage gain is calculated as the product of individual stages or as a sum of all the stages if the gain calculated at each intermediate STAGE was measured in decibels as well.

The correct choice is (b) large

Best explanation: The Small Signal Amplifier is generally referred to as a “Voltage” amplifier because they usually convert a small input voltage into a much LARGER OUTPUT voltage. Although the AMPLIFICATION is high the efficiency of the conversion from the DC power supply input to the AC voltage signal output is usually poor.

The correct choice is (c) 25%

To explain I would say: The theoretical maximum efficiency of a CLASS A POWER amplifier is 50%. In PRACTICE, with the capacitive coupling and inductive loads (loudspeakers), the efficiency can decrease as LOW as 25%. This means 75% of power drawn by the amplifier from the supply line is wasted.

Correct choice is (b) Ro >> RL

The EXPLANATION is: As we KNOW, a current AMPLIFIER is an amplifier which produces an output current proportional to the signal current, where the proportionality factor is independent of the source or load resistances.

Ideally, current amplifier must have zero value of input RESISTANCE and infinite output resistance.

However it indicates that the input resistance should be very small while output resistance should be very large.

Since,Rs>>Ri; High resistance source drives the current amplifier.

Similarly, as Ro >> RLso the current amplifier is driven by low resistance load.

Correct option is (d) Cut off and breakdown

Easy explanation: Value of drain-source voltage, VDS for breakdown with the INCREASE in negative bias voltage is reduced simply due to the FACT that gate-source voltage, VGS keeps adding to the reverse bias at the junction PRODUCED by CURRENT flow. Thus the current flow remains constant between the cut off and breakdown REGION of JFET.