Shahid Posted July 8 Share Posted July 8 QUESTION: The basic purpose of filter is to?A. Minimize variation in ac signalB. Suppress hannonics in rectified outputC. Remove ripples from the rectified outputD. Stabilize dc output voltage EXPLANATIONA filter circuit is a device that removes the a.c. component of rectifier output and allows only d.c. component to reach the load. QUESTION: Non-inverting amplifier circuits have?A. A very high input impedanceB. A very low input impedanceC. A low output impedanceD. None of the above options is correct EXPLANATIONOperational amplifiers have high input impedance and low output impedance because of the concept of a voltage divider, which is how voltage is divided in a circuit depending on the amount of impedance present in given parts of a circuit. Operational amplifiers are voltage-gain devices. QUESTION: The electron volt is the unit of __________.A. Electric currentB. Electric potentialC. Electric energyD. Electric flux EXPLANATIONAn electron volt is a unit of electric energy equal to the work done on an electron in accelerating it through a potential difference of one volt. Other units of electric energy are erg and Joules. QUESTION: The open-circuit test in a transformer is used to measure?A. Copper lossB. Winding lossC. Total lossD. Core loss EXPLANATIONThe open-circuit test is done to find the core loss of the transformer. It is done by keeping one of the windings open (without load, usually the high voltage winding is open) and applying rated voltage to the other winding (usually the low voltage winding because it is easier to apply rated voltage). QUESTION: Full wave rectifier uses:A. One diodeB. Two diodesC. Three diodesD. Four diodes EXPLANATIONThis is a standard answer. You can make a half-wave rectifier with one diode. For a full-wave bridge rectifier, you will need four diodes. QUESTION: Consider a peak rectifier fed by a 60-Hz sinusoid having a peak value Vp = 100V. Let the load resistance R = 10kΩ . Calculate the fraction of the cycle during which the diode is conducting:A. 1.06 %B. 2.06 %C. 3.18 %D. 4.82 % EXPLANATIONThe correct answer is Option C: 3.18 %The best I can explain: w Δt ~ √(2Vr/Vp)Θ = √(2 X 2/100)Θ = 0.2 rad or 3.18 % of the cycle. QUESTION: Which two or more of the following actions would increase the energy stored in a parallel plate capacitor when a constant potential difference is applied across the plates:A. Decreasing the area of the plates & decreasing the separation between the platesB. Decreasing the area of the plates & increasing the separation between the plates & inserting a dielectric between the platesC. Increasing the area of the plates & decreasing the separation between the plates & inserting a dielectric between the platesD. Increasing the area of the plates & increasing the separation between the plates EXPLANATIONThe energy stored by a parallel plate capacitor depends upon the area of the plates, the distance between the plates, and the presence of dielectric between the plates. The energy stored will increase by increasing the Aaea of the plates, decreasing the distance between them, and by placing a dielectric between the plates. QUESTION: To get a peak load voltage of 40V out of a bridge rectifier, what should be the approximate rms value of secondary voltage?A. 0 VB. 14.4 VC. 28.3 VD. 56.6 V EXPLANATIONFollowing is the solution to this question: QUESTION: A cable 4 km long and of total resistance 1 Ohm carries electric current from a generator producing 100kW at 10,000 Volts. The current in amperes in the cable is:A. 10B. 10,000C. 1000D. 100,000 EXPLANATIONThe explanation for this question is given below: QUESTION: In full wave rectifier with input frequency 50 Hz the ripple in the output is mainly of frequencyA. 25 HzB. 50HzC. 100 HzD. Zero EXPLANATIONIn full-wave rectification, ripple frequency is twice the input frequency whereas for half-wave rectifiers it is equal to the AC supply frequency. QUESTION: In half-wave rectification, the output DC voltage is obtained across the load for ______.A. The positive half cycle of input ACB. The negative half cycle of input ACC. The positive and negative half cycles of input ACD. Either positive or negative half cycle of input EXPLANATIONIn half-wave rectification, the output DC voltage is obtained across the load for either the positive or negative half cycle of input. QUESTION: If peak voltage across a full wave rectifier is 20V then Vrms is:A. 7.07vB. 14.14vC. 16.8vD. 12v EXPLANATIONIn a full-wave rectifier, the relationship between peak voltage and the rms(root mean square) is VP = 2 V rms. V rms here is unknown, peak voltage is 20V, which, when divided by the square root of 2, gives us V rms which is roughly 14.14 V. QUESTION: In full wave rectifier with input frequency 50 Hz, the ripple in the output is mainly of frequency? A. 25 Hz B. 50 Hz C. 100 Hz D. Zero EXPLANATIONThe rectifier rectifies both positive and negative sides of the waveforms, doubling the input frequency and giving 100Hz as the output. QUESTION: Transistors can be used as:A. Half wave rectifierB. Full wave rectifierC. Both D. None of these EXPLANATIONTransistors, if their base-emitter or base-collector regions are connected, can function as either a full or half wave rectifier. QUESTION: In half wave rectification, the output DC voltage is obtained across the load for?A. The positive half cycle of input AC B. The negative half cycle of input ACC. The positive and negative half cycles of input ACD. Either positive or negative half cycle of input AC EXPLANATION A half wave rectifier would rectify either positive or negative, not both, hence the output Voltage would be corresponding to the input AC voltage of one of the two polarities. QUESTION: A diode works in ___ bias for rectification.A. ForwardB. ReverseC. MidD. Positive EXPLANATIONWhen the AC voltage is positive on the cathode side of the diode, the diode allows the current to pass through to the output. But when the AC current reverses direction and becomes negative on the cathode side of the diode, the diode blocks the current so that no voltage appears at the output. Under reverse bias, minority carriers free electrons in the p-type semiconductor and the minority carriers holes in the n-type semiconductor start flowing across the junction. Thus, electric current is produced in the reverse bias diode due to the minority carriers. However, the electric current produced by the minority carriers is very small. So, the minority carrier current in the reverse bias condition is neglected. QUESTION: Maximum efficiency of half wave rectifier is:A. 80.6 % B. 40.6 %C. 70%D. 50% EXPLANATIONHalf-wave rectifiers have a large ripple factor and make use of only one half-cycle of the input. Unlike full-wave rectifiers, they don’t even use transformers, so they have a much lower efficiency (half that of a full-wave rectifier). QUESTION: In a centre tap full wave rectifier if V_{m} is the peak voltage across the secondary of transformer, the maximum voltage across reverse bias isA. V_{m}B. V_{m}/v^{2}C. 2V_{m}D. V_{m}/3 EXPLANATIONUsing the concept of Peak Inverse Voltage, it can be found that if the input peak voltage is V_{m}, the Peak Inverse Voltage is 2 times that, hence 2V_{m} QUESTION: Gamma rays are attracted towardsA. Negative plateB. Positive plateC. No deflectionD. Pass through the medium EXPLANATIONGamma rays have no charge, hence 0 deflection, but also their penetrating power allows them to simply pass through most mediums without blockage. QUESTION: Which of the following is not a type of rectifier? A. Phase wave rectifier B. Full wave C. halfwaveD. none of them EXPLANATIONThe two main types of rectifiers are Half Wave & Full Wave rectifiers. Phase Wave is not a type of rectifier. Hence option A is correct. QUESTION: Half wave voltage multiplier can provide any degree of voltage multiplication by cascading diodes and capacitors. A. any doublerB. any triplerC. any multiplicationD. none of them EXPLANATIONHalf wave voltage multipliers can provide any degree of multiplication of the input voltage by cascading diodes and capacitors. Hence option C is correct. QUESTION: A half-wave rectifier uses:A. One diodeB. Two diodesC. Three diodesD. Four diodes EXPLANATIONThis is a fact to be memorized. A half-wave rectifier circuit employs only one diode to generate a Half-Wave Rectified Output. Hence A is the correct answer. QUESTION: Which of the following rectifier uses a wheatstone bridge to rectify signals?A. Half waveB. Full waveC. BridgeD. All of the above options are correct EXPLANATIONThis is a fact to be memorized. All bridge rectifiers are a type of Wheatstone Bridge circuits. Hence C is the correct answer. QUESTION: Consider a peak rectifier fed by a 60-Hz sinusoid having a peak value V_{p} = 100 V. Let the load resistance R = 10 kΩ. Calculate the fraction of the cycle during which the diode is conducting:A. 1.06 %B. 2.06 %C. 3.18 %D. 4.82 % EXPLANATIONw Δt ~ √(2Vr/Vp) = √(2 X 2/100) = 0.2 rad or 3.18 % of the cycle. QUESTION: The process of converting alternating current to direct current is called?A. Modulation.B. Amplification.C. Oscillation.D. Rectification. EXPLANATIONA rectifier converts alternating current to direct current and this process of converting AC to DC is known as rectification. QUESTION: The ripple factor of half-wave rectifier is?A. 1.21.B. 0.8.C. 0.6.D. 0.4. EXPLANATIONRipple factor determines how well a half-wave rectifier can convert AC voltage to DC voltage. The ripple factor of a half-wave rectifier is 1.21. QUESTION: A full wave rectifier uses load resistor of 1500 Ω. Assume the diodes have Rf= 10 Ω, Rr= ∞. The voltage applied to diode is 30V with a frequency of S0Hz. Calculate the AC power input.A. 358.98mW.B. 275.2mW.C. 145.76 mW.D. 456.78 mW. EXPLANATIONThe AC power input; P_{IN}=I_{RMS}^{2}(R_{F}+R_{r}). I_{RMS}= I_{m}/√2 = V_{m}/(R_{f}+R_{L})√2 = 30/[(1500+10) x 1.414] = 13.5mA. So, P_{IN}= (13.5 x 10^{-3})^{2} x (1500+10) =275.2mW. QUESTION: Which of the following is a type of rectifier?A. Half waveB. Full waveC. BridgeD. All of the above options are correct EXPLANATIONA half-wave rectifier is defined as a type of rectifier that allows only one-half cycle of an AC voltage waveform to pass while blocking the other half cycle. A full-wave rectification rectifies the negative component of the input voltage to a positive voltage, then converts it into DC (pulse current). A bridge rectifier is a type of full-wave rectifier that uses four or more diodes in a bridge circuit configuration to efficiently convert alternating (AC) current to a direct (DC) current. Hence, all of the above options are types of rectifiers. QUESTION: If a half-wave rectifier is used to convert 50Hz AC into DC, then the number of pulses present in the rectifier voltage is:A. 25B. 50C. 100D. 75 EXPLANATIONRectifiers produce uni-directional and pulsating voltage from an AC source. The number of pulses per second in the rectified voltage is the same as that of the input voltage, since the input voltage is 50 Hz then the pulses per second are also 50. QUESTION: Transformer is used in rectification to ________ the supply voltage.A. Step upB. Step downC. EqualizeD. None of these options is correct EXPLANATIONTransformers are used in rectification to step down the supply voltage as during rectification we use a transformer to get AC supply and receive a very high voltage of 240V, which can damage or burn out components like diodes, transistors etc. Therefore, transformers are used to convert the voltage into a lower value to protect the components from damage. QUESTION: A full wave rectifier uses a load resistor of 15000. Assume the diodes have RF=l0 Ohms and Rr= infinite. The voltage applied to the diodes is 30V with a frequency of 50Hz. Calculate the AC power input.A. 358.98 mWB. 275.2 mWC. 145.76 mWD. 456.78 mW EXPLANATIONThe AC power input P_{IN}=I_{RMS}2(RF+Rr). I_{(RMS)}=Im/√2 = Vm/(RF+RL)√2 = 30/(1500+10) x 1.414 = 13.5mA So, P_{IN}=(13.5 x 10-3)2 x (1500+10) = 275.2 mW. QUESTION: The sun gives light at the rate of 1500 W/m^{2} of area perpendicular to the direction of light. Assume the wavelength of light as 5000 Å. Calculate the number of photons/s arriving at 1 m^{2} area at that part of the earth.A. 4.770 x 10^{21}B. 3.770 x 10^{11}C. 3. 770 x 10^{2}D. 3.770 x 10^{22} EXPLANATIONGiven that, I = 1500 W/m² λ = 5000 Å Energy of a particle of light (E), called a photon, is proportional to its frequency (f), by a constant factor (h) which is called the Plank’s constant with a value of 6.626 x 10^{-34} kgm²/s. Giving the equation, E=hv E = hc/λ Speed of light = v = 3x108 m/s, Number of photons received = n = IA/E n = IA/hv n = (1500 x 1)(5000 x 10^{-10}) /(6.626 x 10^{-34} )(3 x 108) n = 3.77 x 10^{21} QUESTION: Most widely used rectifier is:A. half wave rectifierB. full wave rectifierC. bridge rectifierD. none of them EXPLANATIONBridge rectifier is the most commonly used rectifier in electronics. QUESTION: Transistors can be:A. half wave rectifierB. full wave rectifierC. bothD. none EXPLANATIONTransistor can be used as a switch for opening and closing of the circuit. A transistor can also act as an amplifier, but not rectifier Since none of these answers are in the options, the answer is D. QUESTION: A full wave rectifier uses load resistor of 1500 ohm. Assume the diodes have R_{f} = 10 ohm, R_{r} = infinity. The voltage applied to diode is 30V with a frequency of 50Hz. calculate AC power input.A. 358.98 mWB. 275.2 mWC. 145.76 mWD. 456.78 mW EXPLANATIONThe AC P(in)=Irms^2(RF+Rr). Irms=Vm/(Rf+RL)√2 =30/(1500+10) x 1.414 =13.5mA So, P(in)=(13.5 x 10^-3)^2 x (1500+10)=275.2mW. QUESTION: In half-wave rectification, during negative cycle of the wave, the diode is:A. Reverse biasedB. Forward biasedC. Potential barrierD. None of these options are correct EXPLANATIONDuring the positive half cycle of the AC voltage, the diode will be forward biased and the current flows through the diode. During the negative half cycle of the AC voltage, the diode will be reverse biased and the flow of current will be blocked. So the correct option is A. QUESTION: The ripple factor of full-wave rectifier is:A. 1.21B. 0.6C. 0.482D. 0.9 EXPLANATIONThe ripple factor for a full-wave rectifier is 0.482 which can also be written as 48.2%. QUESTION: The number of diodes in bridge rectifier is:A. 4B. 3C. 2D. 5 EXPLANATIONBridge rectifiers use four diodes that are arranged cleverly to convert the AC supply voltage to a DC supply voltage. Hence the correct option is A. QUESTION: The ripple factor of a half-wave rectifier is:A. 1.21B. 0.8C. 0.6D. 0.4 EXPLANATIONThe ripple factor of a half-wave rectifier is 1.21. QUESTION: A full-wave rectifier is operating from 50 Hz. The fundamental frequency of the ripple will be:A. 100 HzB. 25 HzC. 50 HzD. 200 Hz EXPLANATIONA full-wave rectifier rectifies both the half cycles of the AC input meaning it conducts twice during a cycle. The output frequency is double to that of the input frequency. The output frequency of a full-wave rectifier is 2 x 50 = 100 Hz. QUESTION: Half-wave rectifier has _____ diodes.A. 1B. 2C. 3D. 4 EXPLANATIONA half-wave rectification is where an alternating voltage of a certain period (input voltage) is applied to a single diode which is connected in series with a load resistance. QUESTION: The number of diodes in a bridge rectifier is:A. 5B. 4C. 3D. 2 EXPLANATIONBridge Rectifiers use four diodes that are arranged cleverly to convert the AC supply voltage to a DC supply voltage. Hence the correct option is A. QUESTION: Rectifier is a device which converts?A. AC to DCB. DC to ACC. AC to triangular currentD. DC to triangular current EXPLANATIONRectifiers convert the current from AC to DC. AC is current that switches between flowing backwards and forwards at regular intervals while DC flows in a single direction. QUESTION: The bridge rectifier is preferred to an ordinary two diode full wave rectifier because:A. It needs much smaller transformer for the same outputB. No center tap requiredC. Less PIV rating per diodeD. All of the above EXPLANATIONFor high voltage applications, a bridge rectifier is preferred because of the low PIV rating per diode. It can be constructed with or without a transformer and it does not require a center tap transformer. QUESTION: Ripple factor is defined as?A. lrms/VrmsB. ldc/lrmsC. lrms/I dcD. lrms +Idc EXPLANATIONThe ripple factor is defined as the ratio of the rms value of the ac component to the dc component. QUESTION: If a half wave rectifier is used to convert 50Hz AC into DC, then the number of pulses present in rectifier voltage is?A. 25B. 50C. 100D. 75 EXPLANATION50 Hz means that current or voltage will change its direction from + to – 50 times in 1 second so it will complete 50 cycles per second. After rectification, the negative half of the cycle will get erased. So for that part of the cycle, the voltage will remain to be Zero. There’s no effect on the cycle length. Hence the number of pulses before and after rectification will remain the same in half-wave rectification. QUESTION: The process of converting alternating current into direct current is called:A. ModulationB. AmplificationC. OssilationD. Rectification EXPLANATIONThe process of converting alternating current and voltage into direct current and voltage is called rectification. This is used in electronic equipment which requires a direct current. For example, voltage in many devices must be rectified from the alternating voltage produced at power stations. QUESTION: In half wave rectification, the output DC voltage is obtained across the load for A. The positive half cycle of input AC B. The negative half cycle of input AC C. The positive and negative half cycles of input AC D. Either positive or negative half cycle of input AC EXPLANATIONThe DC current or DC voltage produced by the negative half wave rectifier is measured across the load resistor RL. The output DC current or DC signal produced by a negative half wave rectifier is a series of negative half cycles or negative sinusoidal pulses. QUESTION: To get a peak load voltage of 40V out of a bridge rectifier, what should be the approximate rms value of secondary voltage? A. 0VB. 14.4V C. 28.3V D. 56.6V EXPLANATIONVm = 40V Vrms = (Vm / √2) = (40 / √2) = 28.28V ∴ To get Vm of 40V at output, Vrms of secondary voltage has to be 28.28V QUESTION: Transformer is used in rectification to ___ the supply voltage.A. Step up B. Step downC. Equalize D. None of them EXPLANATIONIn half wave rectifier, we generally use a step-down transformer because the voltage needed for the diode is very small. Applying a large AC voltage without using transformer will permanently destroy the diode. QUESTION: Most widely used rectifier is ?A. Half wave rectifier B. Full wave rectifier C. Bridge rectifier D. None of them EXPLANATIONThe bridge rectifier consisting of four diodes enables full wave rectification without the need for a centre tapped transformer. The bridge rectifier is an electronic component that is widely used to provide full wave rectification and it is possibly the most widely used circuit for this application. QUESTION: A full wave rectifier passes _____ into positive cycles.A. Lower half cycle B. Upper half cycle C. Both cyclesD. None of these EXPLANATION Full-wave rectification rectifies the negative component of the input voltage to a positive voltage, then converts it into DC (pulse current) utilizing a diode bridge configuration. QUESTION: A half-wave rectifier has ______ diodes.A. 1B. 3C. 2D. 4 EXPLANATIONOnly one diode is needed in the half-wave rectifier. This diode limits the current flow in one direction which means that only half of the AC waveform can pass through the diode. QUESTION: The ripple factor of half-wave rectifier is ___.A. 1.21B. 0.8C. 0.6D. 0.4 EXPLANATIONThe ripple factor of half wave rectifier is equal to 1.21 (this value should be memorized). QUESTION: Which principle is used in solar cells? A. MomentumB. ChargeC. MassD. All of these EXPLANATIONThe light energy is converted to electrical energy in solar cells, while momentum is conserved. QUESTION: Which of the following rectifier uses a wheatstone bridge to rectify the signal:A. Half waveB. Full waveC. BridgeD. All of the above options are correct EXPLANATIONA bridge rectifier is formed by connecting four diodes in the form of a Wheatstone bridge. It also provides full-wave rectification. During the first half of the AC cycle, two diodes are forward biased and during the second half of the AC cycle, the other two diodes become forward biased. QUESTION: Consider a peak rectifier fed by a 60-Hz sinusoid having a peak value Vp = 100 V. Let the load resistance R = 10 kCl. Calculate the fraction of the cycle during which the diode is conducting.A. 1.06 %B. 2.06%C. 3.18%D. 4.82% EXPLANATIONw Δt ~ √(2Vr/Vp) = √(2 X 2/100) = 3.18% of the cycle QUESTION: In a centre tap full wave rectifier, if Vm is the peak voltage across the secondary transformer, the maximum voltage across reverse bias is ____.A. VmB. Vm/√2C. 2VmD. Vm/3 EXPLANATIONThe maximum voltage across reverse bias is twice that of Vm so it is going to be 2Vm. QUESTION: Rectifier is a device which converts A. AC to DCB. DC to ACC. AC to triangular currentD. DC to triangular current EXPLANATIONOption A is correct. Rectifier uses diodes to convert AC to DC by either half wave or full wave rectification. Option B is wrong because that is called inversion. QUESTION: Half wave rectifier passes only A. Lower half cycleB. Upper half cycleC. Both cyclesD. None of them EXPLANATIONOption B is correct. Half wave rectification that only the first half cycle of the sinusoidal wave is passed. Reference; https://www.electronics-tutorials.ws/wp-content/uploads/2013/08/diode39.gif?fit=450%2C153 QUESTION: Ripple factor of full wave rectifier isA. 1.21B. 0.6C. 0.482D. 0.9 EXPLANATIONOption C is correct as seen in the table below: Reference; https://cdn.instrumentationtools.com/wp-content/uploads/2016/07/instrumentationtools.com_half-wave-rectifiers-vs-full-wave-rectifiers.png QUESTION: Half wave voltage multiplier can provide any degree of voltage multiplication by cascading diodes and capacitors. A. any doublerB. any triplerC. any multiplicationD. none of them EXPLANATIONOption C is correct. Answer is in the question. QUESTION: A full wave rectifier passes _____ into positive cycles.A. Lower half cycleB. Upper half cycleC. Both cyclesD. None of them EXPLANATIONBoth cycles are positive since the rectifier turns ac voltage into dc voltage. QUESTION: The maximum efficiency of full wave rectifier is ___.A. 80.60%B. 40.60%C. 70%D. 50% EXPLANATIONIt is a fact that maximum efficiency of a full wave rectifier is 80.60%. QUESTION: A circuit that adds positive or negative de voltage to an input sine wave is called ____.A. ClamperB. ClipperC. Diode clampD. Limiter EXPLANATIONA clamper is an electronic circuit that fixes either the positive or the negative peak excursions of a signal to a defined value by shifting its DC value. QUESTION: A half wave rectifier is operating from 50 Hz mains. The fundamental frequency of ripple will be ___.A. 100 HzB. 25 HzC. 200 HzD. 50 Hz EXPLANATIONAs the output voltage obtained in a half wave rectifier circuit has a single variation in one cycle of ac voltage, hence the fundamental frequency in the ripple of output voltage would be = 50 Hz. QUESTION: The principle behind the working of cathode ray oscilloscope is? A. OscillationB. Half-wave rectificationC. Full wave rectification D. None of these EXPLANATIONOption B is correct. QUESTION: Peak voltage in the output of half-wave rectifier is l0V so de component of the output voltage is? A. 10√ 2B. 10/√ 2C. 10/π D. 10π EXPLANATIONThrough our knowledge, we know the peak voltage is fed with a sinusoidal signal without a filter which is 10V. Here we have a Half wave rectifier, so it nullifies one of the cycles of the input, and only one-half cycle is present in the output. The output DC component of the half-wave rectifier is given by the formula,Vdc = peak voltage (V0) / πVdc = 10/ π V. QUESTION: A full-wave rectifier is operating from 50Hz mains. The fundamental frequency of ripple will be? A. 100 HzB. 25HzC. 50HzD. 200Hz EXPLANATIONFull-wave rectification is a type of rectification in which output is obtained for both halves of a.c. Whereas, the frequency of the output is double that of the input. In a full-wave rectifier, the frequency of the ripple is double that of the main frequency. Hence ripple frequency = 100 Hz. QUESTION: The maximum efficiency of full wave rectifier isA. 80.60%B. 40.60%C. 70%D. 50% EXPLANATIONA full wave rectifier converts AC signals into DC signals. The equation for a full wave rectifier is given by η=Pdc/Pac Where Pdc is the DC power, while Pac is the input power. The max efficiency for a full wave rectifier is 81.2% ~ 80.6%. QUESTION: Average DC voltage across the load in terms of Vmax is:A. 0.532B. 0.637C. 0.759D. 0.437 EXPLANATIONThe average DC voltage across the load is 0.637 Vmax. Because the current flowing through the load is uni-directional, so the voltage developed across the load is also uni-directional, the same as for the previous two diode full-wave rectifier, therefore the average DC voltage across the load is 0.637 Vmax. QUESTION: In a half-wave rectification, during the negative cycle of the wave the diode is A. Reversed biasedB. Forward biasedC. Potential barrierD. None of these EXPLANATIONOption A is correct since,during the negative cycle of the wave, the diode is reverse biased and does not allow that negative half to pass. Option B is incorrect since a diode is forward biased during the positive half of the alternating cycle. Option C is incorrect since the barrier potential is very high so applied voltage cannot exceed that, this is the reason for the reverse biasing of diodes. Option D is incorrect since it states none of these, which is not the case. QUESTION: A half wave rectifier is equivalent to A. Clamper B. ClipperC. Clamper circuit with negative biasD. Clamper circuit with positive bias EXPLANATIONOption B is correct since a clipper works to remove either the positive or negative half cycles of the input voltage. Since the diode most commonly allows the positive half cycles to pass through and not the negative, it acts as a positive clipper. Option A is incorrect since clamping circuits place the negative and positive peaks of the waveform at the desired DC level. In other words, these circuits shift the input signal by an amount defined by the independent voltage source. Option C is incorrect since a negative clamper circuit is biased with some negative reference voltage, that voltage will be added to the output to raise the clamped level. Option D is incorrect since a positive clamper circuit is one that consists of a diode, a resistor, and a capacitor and that shifts the output signal to the positive portion of the input signal. QUESTION: The output voltage of a rectifier is:A. SmoothB. PulsatingC. Perfectly DirectD. Alternating EXPLANATIONWe can use the image below to come to the answer for this question. QUESTION: A full wave rectifier is operating from 50 Hz mains. Fundamental frequency of ripple will be A. 100 HzB. 25 HzC. 50 HzD. 200 Hz EXPLANATIONOption A is correct since the rate of flow of charge is current. Option B is incorrect since the rate of flow of momentum is force. Option C is incorrect since the rate of flow of power is not current. Option D is incorrect since it states none of these, which is not the case. QUESTION: In a half wave rectification,during negative cycle of the wave the diode is?A. Reversed biasedB. Forward biasedC. Potential barrierD. None of these EXPLANATIONIn a half wave rectification, during the positive half cycle the diode is forward biased and during the negative half cycle the diode is reverse biased. QUESTION: Centre tape rectifier circuit consists of ______ diode.A. 1B. 200%C. 300%D. 400% EXPLANATIONA Center tapped full wave rectifier has two diodes while a general full wave rectifier has 4 diodes. QUESTION: Peak voltage in the output of full wave rectifier is 10V so dc component of output voltage is:A. 10√2B. 20/√2C. 20/πD. 20π EXPLANATIONAs vdc = 2 vac / pi vdc = 2(10) / pi vdc = 20 / pi QUESTION: In which rectifier ripple factor is less?A. Full waveB. Half waveC. Both a) and b)D. None of them EXPLANATIONIn a full wave rectifier, lesser ripples are there in the output signal as compared to the half wave rectifier. QUESTION: A half wave rectifier is operating from 50 Hz mains. Fundamental frequency of ripple will be:A. 100 HzB. 25 HzC. 200HzD. 50 Hz EXPLANATIONIn a half-wave rectifier, the ripple frequency of the output remains the same. So, if a half wave rectifier is operating from 50 Hz mains, the fundamental frequency of ripple will be the same i.e., 50Hz. QUESTION: Rectifier is a device which convertsA. AC to DCB. DC to ACC. AC to triangular currentD. DC to triangular current EXPLANATIONRectifier is an electronic device which is used to convert alternating current into direct current. QUESTION: Which of the following rectifier uses wheatstone bridge to rectify signal A. half waveB. full waveC. bridgeD. All of the above EXPLANATIONFull wave rectifiers use a wheatstone bridge arrangement to rectify the signal. QUESTION: In which rectifier ripple factor is lessA. Full waveB. Half waveC. Both A&BD. None of them EXPLANATIONIn a full wave rectifier, the ripple factor is less than the half wave rectifier. QUESTION: For a half-wave rectifier having diode voltage VD and supply input of V, the diode conducts for π-2Θ, where Θ is given by A. tan^{-1}(VD/V)B. sin^{-1}(VD/V)C. cos^{-1}(VD/V)D. cot^{-1}(VD/V) EXPLANATIONThe diode doesn’t conducts when VD ≥VI . Hence Θ = sin^{-1}(D/VI). QUESTION: The principle behind the working of cathode ray oscilloscope is which of the following?A. OscillationB. Half wave rectificationC. Full wave rectificationD. None of these EXPLANATIONOption B is correct since the basic principle behind the working of cathode ray oscilloscope is half wave rectification (fact). QUESTION: Average de Voltage across the load in terms of Vmax is which of the following?A. 0.532 VmaxB. 0.637VmaxC. 0. 759VmaxD. 0.437Vmax EXPLANATIONOption B is correct since the current flowing through the load is unidirectional, so the voltage developed across the load is also unidirectional the same as for the previous two diode full-wave rectifiers, therefore the average DC voltage across the load is 0.637Vmax. QUESTION: Consider a peak rectifier fed by a 60-Hz sinusoid having a peak value Vp = 100 V. Let the load resistance R = 10 kn. Calculate the fraction of the cycle during which the diode is conducting.A. 1.06 %B. 2.06%C. 3.18%D. 4.28% EXPLANATIONOption C is correct since the best explanation is : w Δt ~ √(2Vr/Vp) Θ = √(2 X 2/100) Θ = 0.2 rad or 3.18% of the cycle QUESTION: Why is the bridge rectifier is preferred to an ordinary two diode full wave rectifier?A. it needs much smaller transformer for the same outputB. No center tap requiredC. Less PIV rating per diodeD. All of the above EXPLANATIONOption D is correct since all the above-mentioned points are advantages of a bridge rectifier. QUESTION: In a centre tap full wave rectifier if Vm is the peak voltage across the secondary of transformer, the maximum voltage across reverse bias is ------A. VmB. Vm/√2C. 2VmD. Vm/3 EXPLANATIONOption C is correct since: QUESTION: Half wave rectifier uses _____.A. One diodeB. Two diodesC. Three diodesD. Four diodes EXPLANATIONHalf wave rectifier uses one diode to limit the flow of current in one direction only. This will allow only one half of the AC wave to pass through it. QUESTION: Half wave rectifier passes only _____.A. Lower half cycleB. Upper half cycleC. Both cyclesD. None of them EXPLANATIONThe single-phase half-wave rectifier configuration above passes the positive half of the AC supply waveform with the negative half being eliminated. QUESTION: The bridge rectifier is preferred to an ordinary two diode full wave rectifier because _____.A. It needs much smaller transformer for the same outputB. No center tap requiredC. Less PIV rating per diodeD. All of the above EXPLANATIONFor a bridge rectifier, the Peak inverse voltage is less per diode and the rectifier uses a smaller transformer that does not require a tap. The ordinary two wave full rectifier, in comparison, produces the same output but with larger transformers and have higher PIV values per diode. QUESTION: In a rectifier, the larger the value of the shunt capacitor filter, _____.A. Larger the peak-to-peak value of the ripple voltageB. Larger the peak current in the rectifying diodeC. Longer the time the current pulse flows through the diodeD. Smaller the voltage across the load EXPLANATIONC= ΔQ/ΔV = I x t/ΔV If C increases, t increases for constant input power and voltage. Hence the time t for which current flows through the diode increases. QUESTION: If the line frequency is 50 Hz, the output frequency of bridge rectifier is ____.A. 50HzB. 100HzC. 200HzD. 150Hz EXPLANATIONCorrect Option:B This option is correct because in the output of bridge rectifiers one half cycle is repeated,the frequency will be twice as that of input frequency, so f=100Hz. QUESTION: Average DC voltage across the load in terms of Vmax is _____.A. 0.532 VmaxB. 0.637 VmaxC. 0.759 VmaxD. 0.437 Vmax EXPLANATIONThis option is correct because the current flowing through the load is unidirectional, so the voltage developed across is also unidirectional the same as for the previous two diode full-wave rectifiers, therefore the average DC voltage across the load is 0.67 Vmax. QUESTION: The use of a capacitor filter in a rectifier circuit gives satisfactory performance only when the load ____.A. Current is high EXPLANATIONCorrect Option: B (When the current is low a satisfactory performance is achieved) QUESTION: Which of the following is a type of rectifier?A. Half waveB. Full waveC. BridgeD. All of the above EXPLANATIONOption D is correct.Options A,B,C are incorrect because: These are all types of rectifiers. QUESTION: Ripple factor of half wave rectifier is ____.A. 1.21B. 0.8C. 0.6D. 0.4 EXPLANATIONOption A is correct because: The ripple factor of the half wave rectifier is equal to 1.21 (i.e. γ = 1.21). Options B,C,D are incorrect because: Wrong values. QUESTION: Half wave rectifier passes only ____.A. Lower half cycleB. Upper half cycleC. Both cyclesD. None of them EXPLANATIONOption B is correct because: A half-wave rectifier removes the negative half cycle of an AC input and allows only the positive cycles to pass creating a DC flow. Options A, C, and D are incorrect because the negative half cycle gets eliminated. QUESTION: A full wave rectifier is operating from 50 Hz mains. Fundamental frequency of ripple will be ____.A. 100 HzB. 25 HzC. 50 HzD. 200 Hz EXPLANATIONOption A is correct because: In a full wave rectifier, we get the output for the positive and negative cycle of input a.c. Hence the frequency of the ripple of the output is twice than that of input a.c. i.e. 100Hz Options B,C,D are incorrect because: Wrong values QUESTION: Which of the following is not a type of rectifier?A. Phase wave rectifierB. Full waveC. Half waveD. None of these options are correct EXPLANATIONOption A is correct since a phase wave rectifier is a device that rectifies the input AC voltage. Options B and C are incorrect since half and full waves are the types of a rectifier. Option D is incorrect since option A is the right answer. QUESTION: A full wave rectifier is operating from 50 Hz mains. Fundamental frequency of ripple will be ______.A. 100 HzB. 25 HzC. 50 HzD. 200 Hz EXPLANATIONOption A is correct since in a full wave rectifier the frequency of ripple will be double to that of main frequency, so 100Hz. Options B, C, and D are incorrect since they are wrong answers. QUESTION: Rectifier is a device which converts _____.A. AC to DC EXPLANATIONRectifier is a device that is used to convert Alternating Current to a Direct Current. Hence Option A is correct. QUESTION: In a full-wave rectifier with input frequency 50Hz, the ripple in the output is mainly of frequency _____. A. 25HzB. 50HzC. 100HzD. Zero EXPLANATIONThe frequency of the output voltage or current in a full-wave rectifier is double the input frequency. Therefore an input of 50 Hz will generate an output of 100 Hz. Hence Option C is correct. QUESTION: The ripple factor of a full-wave rectifier is ____.A. 1.21B. 0.5C. 0.482D. 0.9 EXPLANATIONThis is a value to memorized. The ripple factor of a full-wave rectifier is 0.482. Hence Option C is correct. QUESTION: The output voltage of a rectifier is ______.A. SmoothB. PulsatingC. Perfectly directD. Alternating EXPLANATIONA rectifier always converts an Alternating Voltage/Current to a pulsating Direct Voltage/Current. Hence Option B is correct. QUESTION: In a center tap full-wave rectifier, if V_{m} is the peak voltage across the secondary coil of the transformer, the maximum voltage across reverse bias is _____. A. V_{m}B. V_{m}/√2C. 2V_{m}D. V_{m}/√3 EXPLANATIONPeak inverse voltage or peak reverse voltage is the maximum voltage a diode can withstand in the reverse bias condition. This voltage will be double the peak voltage. So 2V_{m}. Hence Option C is correct. QUESTION: In a half wave rectification, during negative cycle of the wave the diode is ______.A. Reversed biased EXPLANATIONDuring half-wave rectification of AC, during one half of the cycle, the diode is forward biased while during the other half (negative cycle), it is reversed biased. Hence, Option A is correct. QUESTION: If a half wave rectifier is used to convert 50Hz AC into DC, then the number of pulses present in rectifier voltage is _____.A. 25B. 50C. 100D. 75 EXPLANATIONIf, for example, a 2 Hz (2 cycles per second) input alternating voltage is half-wave rectified, the output signal will have two peaks, one from each cycle, in one second. So if a 50 Hz alternating voltage is half-wave rectified, the output will have 50 pulses/peaks. Hence, Option B is correct. QUESTION: The number of diodes in bridge rectifier is _____.A. 4B. 3C. 2D. 5 EXPLANATIONA full-wave bridge rectifier circuit uses four diodes. Hence, Option A is correct. QUESTION: Input resistance of an Operational Amplifier is:A. Very lowB. Very highC. ModerateD. Becomes infinite EXPLANATIONThe input resistance is infinite for an ideal OP-AMP which prevents current from flowing into the OP-AMP itself as it may damage the component. However, since we are not talking about an ideal situation, in real life the resistance is not infinite rather very high thus the answer is B. QUESTION: Operational amplifier with an LDR is used as a:A. Half wave rectifierB. GateC. Night switchD. Full wave rectifier EXPLANATIONC is the answer as the polarity of the output of the OP-AMP would change based on the input and since the LDR is placed on one of the inputs, we can say that the output polarity is dependent on the light intensity which is perfect for a night switch. QUESTION: A capacitor and an inductor are connected to A.C. supply. If we double the frequency of A.C. supply then the effect on reactance of capacitor X_{C} and reactance of inductor X_{L} will be:A. Both X_{C} and X_{L} are doubled B. Both X_{C} and X_{L} are halved C. X_{C} is doubled and X_{L} reduces to half D. X_{L} is doubled and X_{C} reduces to half E. Both X_{C} and X_{L} are unchanged EXPLANATIONThe inductive reactance of an inductor increases as the frequency across it increases therefore inductive reactance is proportional to frequency ( X_{L} α ƒ ) The capacitive reactance of the capacitor decreases as the frequency across it increases therefore capacitive reactance is inversely proportional to frequency. QUESTION: When an Operational amplifier is used as Inverting amplifier, then angle between its output and input signal is:A. 30 degrees B. 45 degrees C. 60 degrees D. 90 degrees E. 180 degrees EXPLANATIONIt is called Inverting Amplifier because the op-amp changes the phase angle of the output signal exactly 180 degrees out of phase with respect to input signal. QUESTION: A capacitor and an inductor are connected to A.C. supply. If we double the frequency of A.C. supply then the effect on reactance of capacitor X_{C} and reactance of inductor X_{L} will be:A. Both X_{C} and X_{L} are doubled B. Both X_{C} and X_{L} are halved C. X_{C} is doubled and X_{L} reduces to half D. X_{L} is doubled and X_{C} reduces to half E. Both X_{C} and X_{L} are unchanged EXPLANATIONThe inductive reactance of an inductor increases as the frequency across it increases therefore inductive reactance is proportional to frequency ( X_{L} α ƒ ) The capacitive reactance of the capacitor decreases as the frequency across it increases therefore capacitive reactance is inversely proportional to frequency. QUESTION: When an Operational amplifier is used as Inverting amplifier, then angle between its output and input signal is:A. 30 degrees B. 45 degrees C. 60 degrees D. 90 degrees E. 180 degrees EXPLANATIONIt is called Inverting Amplifier because the op-amp changes the phase angle of the output signal exactly 180 degrees out of phase with respect to input signal. QUESTION: When reverse bias of a p-n junction diode is increased, a stage comes when electric field set up is sufficient to rupture covalent bonds and release a large number of holes and electrons. This breakdown phenomenon is known as:A. Avalanche breakdownB. Zoner breakdownC. Early effectD. High - field emission EXPLANATIONAvalanche breakdown (or “the avalanche effect”) is a phenomenon that can occur in both insulating and semiconducting materials. It is a form of electric current multiplication that can allow very large currents within materials which are otherwise good insulators. It is a type of electron avalanche. QUESTION: If both inputs “A” and “B” are ON and the output is also ON. This type of gate is known as:A. OR GateB. AND GateC. NAND GateD. NOR Gate EXPLANATIONTechnically, the answer can either be A or B here because putting ON at both inputs will give ON in both, as an output. But we will choose B here, since this condition is specific here. Only ON and ON give ON in AND, while OR gate can give ON with other combinations too. QUESTION: ______ is equivalent to AND Gate followed by a NOT Gate.A. NOR GateB. NAND GateC. NOT GateD. None of these EXPLANATIONWhen the AND and NOT gate are combined it forms a NAND gate. Outputs are opposite to the AND gate. QUESTION: Which option in the following best explains Base of a transistor?A. Base has no impurityB. Base has less concentration of impurity as compared to emitter and collectorC. Base has more concentration of impurity as compared to emitterD. Base has more concentration of impurity as compared to collector EXPLANATIONThis is also the reason that base is very lightly doped because its only purpose is to act as a control for the collector current QUESTION: A logic gate has four inputs. Its possible input combinations will be:A. 4B. 16C. 32D. 64 EXPLANATIONThis can be calculated using the formula 2^{n}, where n is the number of possible inputs. The number of inputs is 4. the each input will take 0 OR 1 as values Therefore 2^{4} = 16 possible combinations of inputs are there as QUESTION: In a npn transistor, the current IE that flows in the emitter circuit is: A. I_{C} + I_{B} B. I_{C} - I_{B} C. I_{C} x I_{B} D. Zero EXPLANATIONEmitting current=Collecting current+Base current. QUESTION: In photodiodes, the bias voltage is applied in ______ bias form. A. Forward B. Resistance C. Reverse D. Converse EXPLANATIONThe photodiode is reverse biased for operating in the photoconductive mode. As the photodiode is in reverse bias the width of the depletion layer increases. This reduces the junction capacitance and thereby the response time. In effect, the reverse bias causes faster response times for the photodiode. QUESTION: A semiconductor photodiode is a: A. reverse biased junction diode B. forward biased junction diode C. half wave rectifier D. full wave rectifier E. Transistor EXPLANATIONThe photodiode is reverse biased for operating in the photoconductive mode. As the photodiode is in reverse bias the width of the depletion layer increases. This reduces the junction capacitance and thereby the response time. In effect, the reverse bias causes faster response times for the photodiode. QUESTION: A negligible small current between input terminals of the operational amplifier is because of:A. High output resistanceB. Low output resistanceC. High input resistanceD. Low input resistance EXPLANATIONIdeal op-amps have zero input current. A small input current passes because of very high input impedence. QUESTION: A signal of -80 mV is applied to the inverting terminal of the amplifier while the non inverting terminal is grounded. The gain of the amplifier is 25 using Rin(R1) equal to 3 ohms and Rf(R2) equal to 75 Ohms. What would be the value of the output signal?A. 200 mVB. -3 VC. 2 VD. 3 V EXPLANATIONSince Gain = 25 Vout = Vin * Gain = -80mv * 25 = -2000 mv = -2V = 2V** ** since a positive-going voltage at the inverting terminal gives a negative-going output voltage, hence the answer is positive 2V. QUESTION: When we apply input at the non-inverting input of an operational amplifier then the output appears after:A. Amplification, phase shift of 180˚B. Amplification, phase shift of 60˚C. Amplification, no phase shiftD. None of the given options EXPLANATIONThe voltage gain is always greater than one. The voltage gain is positive, indicating that for AC input, the output is in-phase with the input signal and amplified QUESTION: Which of the following is false?A. In P-type substance majority charge carriers are holesB. During forward biasing of a divide width of depletion region decreasesC. Leakage current flows to minority charge carriersD. None of the given options EXPLANATIONAll the options are true QUESTION: Which of the following is not true about LEDs?A. There are two wires (called leads) attached to an LEDB. While connecting to a circuit, the longer lead of LED is always connected to the positive terminal of the batteryC. While connecting to a circuit, the shorter lead of LED is always connected to the positive terminal of the batteryD. LED can be used to detect weak current EXPLANATIONWhile connecting to a circuit, the longer lead is always connected to the positive terminal of the battery and the shorter lead is connected to the negative terminal of the battery. QUESTION: Which of the following is the truth table for the logic gate: A. AB. BC. CD. D EXPLANATIONExplanation will be added soon QUESTION: If the fundamental logic gates are connected as: A. AB. BC. CD. D EXPLANATIONExplanation will be added soon QUESTION: Three NAND gates are connected as shown in the figure.Which of the following logic gates is formed in the connected circuit? A. ORB. ANDC. NORD. NAND EXPLANATIONWe can check to which gate this diagram refers to by sending in 1 from A and 0 from B.Output at X, after working, will be 1. This means that this gate is working as an OR gate. Asonly in OR gate, when two inputs are 1 and 0 respectively the output is 1. QUESTION: What is the output of the truth table? A. A B. A AND CC. BD. NO CORRECT OPTION EXPLANATIONA.B means that AAND B imply that 0 AND 0 should give 0,1 and 0 should give 0 and 1 and1 should give 1.This makes the equation for four Xs as0 OR 0=00 OR 0=00 OR 0=01 OR 1=1 QUESTION: The difference between the plates of a parallel plate capacitor is 2.0 mm and area ofeach plate is 2.0 m^2. The plates are in a vacuum. A potential difference of 1.0 x10^4 V is applied across the plates. Find the capacitance. A. 4 x 10-3 F B. 3.54 x 10-9 F C. 8.85 x 10-9 F D. 9.0 x 10-9 F EXPLANATIONThe capacitance of a parallel plate capacitor is C=ϵ0AdC=(8.85x10^-12)*(2)*(2x10^-3)=3.54 x 10^-9C QUESTION: Which of the following techniques is the practical application of X-rays? A. Magnetic Resonance Imaging B. Ultrasonography C. Computerized Axial Topography D. Positron Emission Tomography EXPLANATIONComputed tomography (CT) scanning, also known as, especially in the olderliterature and textbooks, computerized axial tomography (CAT) scanning, is adiagnostic imaging procedure that uses x-rays to build cross-sectional images(slices) of the body. QUESTION: Which one of the following spectra is most typical of the output of an X-ray tube? A. AB. BC. CD. D EXPLANATIONThe most typical outcome of an X-ray tube is shown in B, which is a factual recall from thebook QUESTION: Which one of the following has the largest energy content? A. γ-rays B. X-rays C. Infra-red radiations D. Ultra-violet radiations EXPLANATIONGamma rays have the highest energy in all of the above mentioned, it is because of theirhighest frequency among all, as shown below: QUESTION: What will be the energy of the accelerated electron used to produce X-rays when theaccelerating potential is 2 kV? A. 2 x 10-19 J B. 1.6 x 10-19 J C. 3.2 x 1019 J D. 3.2 x 10-16 J EXPLANATIONThe formula relating energy, charges, and voltages is:ΔPE = qΔV.PE=(2x10^3)*(1.6x10^-19)PE=3.2x10^-16 J. QUESTION: Process of generating three dimensional images of objects by using laser beam is called A. Photography B. 3-D cinema C. Holography D. Tomography EXPLANATIONHolography is a process that creates three-dimensional images called hologramsusing laser beams, the properties of interference and diffraction, light intensityrecording, and illumination of the recording. QUESTION: What is the output Boolean expression of logic diagram shown in figure below: A. AB. BC. CD. D EXPLANATIONThe outputs of each NAND gate is given by AB and AB, making them the 2 inputs for the OR gate, so the final output X would be AB+AB. QUESTION: Which of the following is the proper way to study the sinusoidal wave form of voltage? A. Voltage is connected to ‘Y’ input and time base is switched onB. Voltage is connected to ‘X’ input and time base is switched offC. Voltage is connected to ‘Y’ input and time base is switched offD. Voltage is connected to ‘X’ input and time base is switched on EXPLANATIONVoltage should be connected to Y and timebase switched on. QUESTION: What is the logic symbol for a NOT Gate?A. Option AB. Option BC. Option CD. Option D EXPLANATIONIn the following diagrams, A is a NOT gate. B is a OR gate, C is AND gate and D is a symbol for PN junction diode. QUESTION: Which one of the following is I-V curve of a junction diode?A. Option AB. Option BC. Option CD. Option D EXPLANATIONThe primary function of a semiconductor diode is rectification of AC to DC. When a diode is forward biased ; the higher potential is connected to its Anode, it will pass current. When the diode is reverse biased ; the higher potential is connected to its Cathode, the current is blocked.Then a PN junction needs a bias voltage of a certain polarity and amplitude for current to flow. This bias voltage also controls the resistance of the junction and therefore the flow of current through it.This can further be explained by the diagram below QUESTION: Which of the following rectifier uses wheatstone bridge to rectify signal:A. Half waveB. Full waveC. BridgeD. Both half and full wave EXPLANATIONBridge rectifier is formed by connecting four diodes in the form of a Wheatstone bridge. It also provides full wave rectification. QUESTION: A full wave rectifier passes _________ into positive cycles:A. Lower half cycleB. Upper half cycleC. Both lower and upper half cyclesD. One quarter cycle EXPLANATIONRefer to the following diagram of full wave rectification. QUESTION: In the case of germanium, the value of potential barrier develops across the depletion region isA. 0vB. 0.3VC. 0.7VD. 0.9V EXPLANATIONTypically at room temperature the voltage across the depletion layer for germanium is about 0.3 – 0.35 volts. 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