McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Electrical Engineering .
| 151. |
Given that capacitance w.r.t the input node is 2pF and output node is 4pF, find capacitance between input and output node. |
| A. | 0.67 pf |
| B. | 1.34pf |
| C. | 0.44pf |
| D. | 2.2pf |
| Answer» A. 0.67 pf | |
| Explanation: c1=c(1-k), c2=c(1-k-1) c1=2pf | |
| 152. |
Consider an RC coupled amplifier at low frequency. Internal voltage gain is -120. Find the voltage gain magnitude, when given that collector resistance = 1kΩ, load = 9kΩ, collector capacitance is 0. is 0.1μF, and input frequency is 20Hz. |
| A. | 120 |
| B. | 12 |
| C. | 15 |
| D. | -12 |
| Answer» C. 15 | |
| Explanation: av = -120 | |
| 153. |
15Hz |
| A. | 22.73 hz |
| B. | 612 hz |
| C. | 673hz |
| D. | 317 hz |
| Answer» B. 612 hz | |
| Explanation: rc = 2kΩ, rl = 5kΩ, cc = 1μf, cb = 10μf, ce = 20μf, rs = 2 kΩ | |
| 154. |
Find the 3-dB frequency given that the gain of RC coupled amplifier is 150, the low frequency voltage gain is 100 and the input frequency is 50Hz. |
| A. | 50.8 hz |
| B. | 55.9 hz |
| C. | 60hz |
| D. | 100hz |
| Answer» B. 55.9 hz | |
| Explanation: avm = 150 avl = 100 | |
| 155. |
What is the phase shift in RC coupled CE amplifier at lower 3dB frequency? |
| A. | 180° |
| B. | 225° |
| C. | 270° |
| D. | 100° |
| Answer» B. 225° | |
| Explanation: total phase shift = 180°+ tan- 1(fl/f) | |
| 156. |
We cannot use h-parameter model in high frequency analysis because |
| A. | they all can be ignored for high frequencies |
| B. | junction capacitances are not included in it |
| C. | junction capacitances have to be included in it |
| D. | ac analysis is difficult for high frequency using it |
| Answer» B. junction capacitances are not included in it | |
| Explanation: the effect of smaller capacitors is considerable in high frequency analysis of analog circuits, and hence they cannot be ignored as. instead of h-parameter model, we use π-model. | |
| 157. |
Which of the statement is incorrect? |
| A. | at unity gain frequency the ce short circuit current gain becomes 1 |
| B. | unity gain frequency is the same as gain bandwidth product of bjt |
| C. | gain of bjt decreases at higher frequencies due to junction capacitances |
| D. | β- cut-off frequency is one where the ce short circuit current gain becomes β/2 |
| Answer» A. at unity gain frequency the ce short circuit current gain becomes 1 | |
| Explanation: gain bandwidth product for any mosfet is ft = gm/2π(cgs+cgd) thus gbp is approximately 5.9 mhz. | |
| 158. |
The collector current will not reach the steady state value instantaneously because of |
| A. | stray capacitances |
| B. | resistances |
| C. | input blocking capacitances |
| D. | coupling capacitance |
| Answer» A. stray capacitances | |
| Explanation: when a pulse is given, the collector current will not reach the steady state value instantaneously because of stray capacitances. the charging and discharging of capacitance makes the current to reach a steady state value after a given time constant. | |
| 159. |
For the BJT, β=∞, VBEon=0.7V VCEsat=0.7V. The switch is initially closed. At t=0, it is opened. At which time the BJT leaves the active region? |
| A. | 20ms |
| B. | 50ms |
| C. | 60ms |
| D. | 70ms |
| Answer» B. 50ms | |
| Explanation: at t < 0, the bjt is off in cut off region. ib=0 as β=∞, so ic=ie. when t > 0, switch opens and bjt is on. the voltage across capacitor increases. from the input loop, -5-vbe-i(4.3k)+10=0 and gives i=1ma. ic1=1-0.5=0.5ma. vc1=0.7+4.3+10=-5v. ic1=c1dvc1/dt. from | |
| 160. |
The technique used to quickly switch off a transistor is by |
| A. | reverse biasing its emitter to collector junction |
| B. | reverse biasing its base to collector junction |
| C. | reverse biasing its base to emitter junction |
| D. | reverse biasing any junction |
| Answer» C. reverse biasing its base to emitter junction | |
| Explanation: the technique used to quickly switch off a transistor is by reverse biasing its base to collector junction. it is demonstrated in a high voltage switching circuit. the advantage of this circuit is that it is not necessary to have high voltage control signal. | |
| 161. |
Which of the following circuits helps in the applications of switching times? |
| A. | c) |
| B. | d) |
| Answer» B. d) | |
| Explanation: this is an inverter, in which the transistor in the circuit is switched between cut off and saturation. the load, for example, can be a motor or a light emitting diode or any other electrical device. | |
| 162. |
Which of the following helps in reducing the switching time of a transistor? |
| A. | a resistor connected from base to ground |
| B. | a resistor connected from emitter to ground |
| C. | a capacitor connected from base to ground |
| D. | a capacitor connected from emitter to ground |
| Answer» A. a resistor connected from base to ground | |
| Explanation: connecting a resistor connected from base of a transistor to ground/negative voltage helps in reducing the switching the switching time of the transistor. when transistor saturate, there is stored charge in the base that must be removed before it turns off. | |
| 163. |
The time taken for a transistor to turn from saturation to cut off is |
| A. | inversely proportional to charge carriers |
| B. | directly proportional to charge carriers |
| C. | charging time of the capacitor |
| D. | discharging time of the capacitor |
| Answer» B. directly proportional to charge carriers | |
| Explanation: when sufficient charge carriers exist, the transistor goes into saturation. | |
| 164. |
Switching speed of P+ junction depends on |
| A. | mobility of minority carriers in p junction |
| B. | life time of minority carriers in p junction |
| C. | mobility of majority carriers in n junction |
| D. | life time of minority carriers in n junction |
| Answer» D. life time of minority carriers in n junction | |
| Explanation: switching leads to move holes in p region to n region as minority carriers. removal of this accumulation determines switching speed. p+ regards to a diode in which the p type is doped excessively. | |
| 165. |
Fixed voltage regulators and adjustable regulators are often called as |
| A. | series dissipative regulators |
| B. | shunt dissipative regulators |
| C. | stray dissipative regulators |
| D. | all the mentioned |
| Answer» A. series dissipative regulators | |
| Explanation: series dissipative regulators simulate a variable resistance between the input voltage & the load and hence functions in a linear mode. | |
| 166. |
Linear series regulators are suited for application with |
| A. | high current |
| B. | medium current |
| C. | low current |
| D. | none of the mentioned |
| Answer» B. medium current | |
| Explanation: in series dissipative regulator, conversion efficiency decreases as the input or output voltage differential increases and vice versa. so, linear series regulators are suited for medium current application with a small voltage differential. | |
| 167. |
A series switching regulators |
| A. | improves the efficiency of regulators |
| B. | improves the flexibility of switching |
| C. | enhance the response of regulators |
| D. | all of the mentioned |
| Answer» A. improves the efficiency of regulators | |
| Explanation: a series switching regulators is constructed such that, a series pass transistor is used as a switch rather than as a variable resistance in linear mode. | |
| 168. |
The switching regulators can operate in |
| A. | step up |
| B. | step down |
| C. | polarity inverting |
| D. | all the mentioned |
| Answer» D. all the mentioned | |
| Explanation: the switching regulators can operate in any one of the three modes depending on the way in which the components are connected. | |
| 169. |
It may be required to store energy for a specified amount of time during power failure especially if the system is designed for a computer power supply. |
| A. | 1 and 3 |
| B. | 1,2,3 and 4 |
| C. | 2,3 and 4 |
| D. | 1,3 and 4 |
| Answer» B. 1,2,3 and 4 | |
| Explanation: a voltage source must satisfy the entire given requirement to be used in switching regulator. | |
| 170. |
Which among the following act as a switch in switching regulator? |
| A. | rectifiers |
| B. | diode |
| C. | transistors |
| D. | relays |
| Answer» C. transistors | |
| Explanation: a transistor is connected as power switch and is operated in the saturated mode.thus, the pulse generator output alternatively turns the switch on and off in switching regulator. | |
| 171. |
What should be the frequency range of pulse generator? |
| A. | 250 khz |
| B. | 40 khz |
| C. | 120 khz |
| D. | 20 khz |
| Answer» D. 20 khz | |
| Explanation: the most effective frequency range for pulse generator for optimum efficiency and component size is 20khz. | |
| 172. |
Filter used in switching regulator’s are also as called |
| A. | dc – ac transformers |
| B. | ac – dc transformers |
| C. | dc transformer |
| D. | ac transformer |
| Answer» C. dc transformer | |
| Explanation: filter converts the pulse waveform from the output of the switch into a dc voltage. since this switching mechanism allows a conversion similar to transformers, the switching regulators is often referred to as a dc transformer. | |
| 173. |
Which of the following is considered to be the most important components of the switching regulator? |
| A. | rc or rlc filter |
| B. | rl or rlc filter |
| C. | orc or rl filter |
| D. | rc, rlc or rl filter |
| Answer» B. rl or rlc filter | |
| Explanation: rl or rlc filter is the most important components of the switching regulator, because there are several areas that are affected by the choke of inductor including energy storage for the regulators output ripple, transient, response etc. | |
| 174. |
Which is the most commonly used low voltage switching regulators? |
| A. | powdered permalloy toroids |
| B. | fermite ei, u and toroid cores |
| C. | silicon steel ei butt stacks |
| D. | none of the mentioned |
| Answer» C. silicon steel ei butt stacks | |
| Explanation: the silicon steel ei butt stack exhibits high permeability high flux density and ease of construction and mounting therefore, it is most commonly used in low voltage switching regulators. | |
| 175. |
The diode in a half wave rectifier has a forward resistance RF. The voltage is Vmsinωt and the load resistance is RL. The DC current is given by |
| A. | vm/√2rl |
| B. | vm/(rf+rl)π |
| C. | 2vm/√π |
| D. | vm/rl |
| Answer» B. vm/(rf+rl)π | |
| Explanation: for a half wave rectifier, the idc=iavg=im/π | |
| 176. |
What is the output as a function of the input voltage (for positive values) for the given figure. Assume it’s an ideal op-amp with zero forward drop (Di=0) |
| A. | 0 |
| B. | -vi |
| C. | vi |
| D. | 2vi |
| Answer» C. vi | |
| Explanation: when the input of the inverted mode op-amp is positive, the output is negative. | |
| 177. |
In a half wave rectifier, the sine wave input is 50sin50t. If the load resistance is of 1K, then average DC power output will be? |
| A. | 3.99v |
| B. | 2.5v |
| C. | 5.97v |
| D. | 6.77v |
| Answer» B. 2.5v | |
| Explanation: the standard form of a sine wave is vmsinωt. by comparing the given | |
| 178. |
Efficiency of a half wave rectifier is |
| A. | 50% |
| B. | 60% |
| C. | 40.6% |
| D. | 46% |
| Answer» C. 40.6% | |
| Explanation: efficiency of a rectifier is the effectiveness to convert ac to dc. for half wave it’s 40.6%. it’s given by, vout/vin*100. | |
| 179. |
If peak voltage for a half wave rectifier circuit is 5V and diode cut in voltage is 0.7, then peak inverse voltage on diode will be? |
| A. | 5v |
| B. | 4.9v |
| C. | 4.3v |
| D. | 6.7v |
| Answer» C. 4.3v | |
| Explanation: piv is the maximum reverse bias voltage that can be appeared across a diode in the given circuit, if the piv rating is less than this value of breakdown of diode will occur. for a rectifier, piv=vm-vd=5- 0.7=4.3v. | |
| 180. |
Transformer utilisation factor of a half wave rectifier is |
| A. | 0.234 |
| B. | 0.279 |
| C. | 0.287 |
| D. | 0.453 |
| Answer» C. 0.287 | |
| Explanation: transformer utilisation factor is the ratio of ac power delivered to load to the dc power rating. this factor indicates effectiveness of transformer usage by rectifier. for a half wave rectifier, it’s low and equal to 0.287. | |
| 181. |
If the input frequency of a half wave rectifier is 100Hz, then the ripple frequency will be |
| A. | 150hz |
| B. | 200hz |
| C. | 100hz |
| D. | 300hz |
| Answer» C. 100hz | |
| Explanation: the ripple frequency of the output and input is same. this is because, one half cycle of input is passed and other half cycle is seized. so, effectively the frequency is the same. | |
| 182. |
Efficiency of a centre tapped full wave rectifier is |
| A. | 50% |
| B. | 46% |
| C. | 70% |
| D. | 81.2% |
| Answer» D. 81.2% | |
| Explanation: efficiency of a rectifier is the effectiveness to convert ac to dc. it’s obtained by taking ratio of dc power output to maximum ac power delivered to load. it’s usually expressed in percentage. for centre tapped full wave rectifier, it’s 81.2%. | |
| 183. |
A full wave rectifier delivers 50W to a load of 200Ω. If the ripple factor is 2%, calculate the AC ripple across the load. |
| A. | 2v |
| B. | 5v |
| C. | 4v |
| D. | 1v |
| Answer» A. 2v | |
| Explanation: we know that, p =v 2/r . | |
| 184. |
In a centre tapped full wave rectifier, RL=1KΩ and for diode Rf=10Ω. The primary voltage is 800sinωt with transformer turns ratio=2. The ripple factor will be |
| A. | 54% |
| B. | 48% |
| C. | 26% |
| D. | 81% |
| Answer» B. 48% | |
| Explanation: the ripple factor ϒ= [(irms/iavg)2 – 1]1/2. irms =im / | |
| 185. |
Due to centre tap Vm=400/2=200) IRMS=198/√2=140mA, IAVG=2*198/ π=126mA. ϒ=[(140/126)2-1]1/2=0.48. So, ϒ=48%. |
| A. | 0.623 |
| B. | 0.678 |
| C. | 0.693 |
| D. | 0.625 |
| Answer» C. 0.693 | |
| Explanation: transformer utilisation factor is the ratio of ac power delivered to load to the dc power rating. this factor indicates effectiveness of transformer usage by rectifier. for a half wave rectifier, it’s low and equal to 0.693. | |
| 186. |
If the peak voltage on a centre tapped full wave rectifier circuit is 5V and diode cut in voltage is 0.7. The peak inverse voltage on diode is |
| A. | 4.3v |
| B. | 9.3v |
| C. | 5.7v |
| D. | 10.7v |
| Answer» B. 9.3v | |
| Explanation: piv is the maximum reverse bias voltage that can be appeared across a | |
| 187. |
What is done in switching regulators to minimize its power dissipation during switching? |
| A. | uses external transistor |
| B. | uses 1mh choke |
| C. | uses external transistor and 1mh choke |
| D. | none of the mentioned |
| Answer» C. uses external transistor and 1mh choke | |
| Explanation: to minimize power dissipation during switching, the external transistor must be a switching power transistor and a 1mh choke smooth out the current pulses delivered to the load. | |
| 188. |
33/ Ipk |
| A. | 1- iii , 2- i , 3- ii |
| B. | 1- i , 2- ii , 3- iii |
| C. | 1- iii , 2- ii , 3- i |
| D. | 1- iii , 2- ii , 3- i |
| Answer» B. 1- i , 2- ii , 3- iii | |
| Explanation: characteristics and design formula for step up, step down and converting mode of switching regulator. | |
| 189. |
Choose the correct statement |
| A. | mosfet is a unipolar, voltage controlled, two terminal device |
| B. | mosfet is a bipolar, current controlled, three terminal device |
| C. | mosfet is a unipolar, voltage controlled, three terminal device |
| D. | mosfet is a bipolar, current controlled, two terminal device |
| Answer» C. mosfet is a unipolar, voltage controlled, three terminal device | |
| Explanation: mosfet is a three terminal device, gate, source & drain. it is voltage controlled unlike the bjt & only electron current flows. | |
| 190. |
The arrow on the symbol of MOSFET indicates |
| A. | that it is a n-channel mosfet |
| B. | the direction of electrons |
| C. | the direction of conventional current flow |
| D. | that it is a p-channel mosfet |
| Answer» B. the direction of electrons | |
| Explanation: the arrow is to indicate the direction of electrons (opposite to the direction of conventional current flow). | |
| 191. |
The controlling parameter in MOSFET is |
| A. | vds |
| B. | ig |
| C. | vgs |
| D. | is |
| Answer» B. ig | |
| Explanation: the gate to source voltage is the controlling parameter in a mosfet. | |
| 192. |
In the internal structure of a MOSFET, a parasitic BJT exists between the |
| A. | source & gate terminals |
| B. | source & drain terminals |
| C. | drain & gate terminals |
| D. | there is no parasitic bjt in mosfet |
| Answer» B. source & drain terminals | |
| Explanation: examine the internal structure of a mosfet, notice the n-p-n structure between the drain & source. a p-channel mosfet will have a p-n-p structure. | |
| 193. |
In the transfer characteristics of a MOSFET, the threshold voltage is the measure of the |
| A. | minimum voltage to induce a n-channel/p- channel for conduction |
| B. | minimum voltage till which temperature is constant |
| C. | minimum voltage to turn off the device |
| D. | none of the above mentioned is true |
| Answer» A. minimum voltage to induce a n-channel/p- channel for conduction | |
| Explanation: it is the minimum voltage to induce a n-channel/p-channel which will allow the device to conduct electrically through its length. | |
| 194. |
The output characteristics of a MOSFET, is a plot of |
| A. | id as a function of vgs with vds as a parameter |
| B. | id as a function of vds with vgs as a parameter |
| C. | ig as a function of vgs with vds as a parameter |
| D. | ig as a function of vds with vgs as a parameter |
| Answer» B. id as a function of vds with vgs as a parameter | |
| Explanation: it is id vs vds which are plotted for different values of vgs (gate to source voltage). | |
| 195. |
SMPS is used for |
| A. | obtaining controlled ac power supply |
| B. | obtaining controlled dc power supply |
| C. | storage of dc power |
| D. | switch from one source to another |
| Answer» B. obtaining controlled dc power supply | |
| Explanation: smps (switching mode power supply) is used for obtaining controlled dc power supply. | |
| 196. |
Choose the incorrect statement. |
| A. | smps is less sensitive to input voltage variations |
| B. | smps is smaller as compared to rectifiers |
| C. | smps has low input ripple |
| D. | smps is a source of radio interference |
| Answer» C. smps has low input ripple | |
| Explanation: smps has higher output ripple and its regulation is worse. | |
| 197. |
is used for critical loads where temporary power failure can cause a great deal of inconvenience. |
| A. | smps |
| B. | ups |
| C. | mps |
| D. | rccb |
| Answer» B. ups | |
| Explanation: uninterruptible power supply is used where loads where temporary power failure can cause a great deal of inconvenience. | |
| 198. |
is used in the rotating type UPS system to supply the mains. |
| A. | dc motor |
| B. | self excited dc generator |
| C. | alternator |
| D. | battery bank |
| Answer» C. alternator | |
| Explanation: when the supply is gone, the diesel engine is started, which runs the alternator and the alternator supplies power to the mains. non-rotating type ups are not used anymore. | |
| 199. |
Static UPS requires |
| A. | only rectifier |
| B. | only inverter |
| C. | both inverter and rectifier |
| D. | none of the mentioned |
| Answer» C. both inverter and rectifier | |
| Explanation: rectifier to converter the dc from the battery to ac. inverter to charge the battery from mains. | |
| 200. |
Usually batteries are used in the UPS systems. |
| A. | nc |
| B. | li-on |
| C. | lead acid |
| D. | all of the mentioned |
| Answer» C. lead acid | |
| Explanation: lead acid batteries are cheaper and have certain advantages over the other types. nc batteries would however be the best, but are three to four times more expensive than lead acid. | |
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