Electronic Circuits I Set 7

Electrical Engineering

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This set of Electronic Circuits 1 Multiple Choice Questions & Answers (MCQs) focuses on Electronic Circuits I Set 7

Q1 | In an RC coupled CE amplifier, when the input frequency increases, which of these are incorrect?
  • reactance csh decreases
  • voltage gain increases
  • voltage gain decreases due to shunt capacitance
  • an rc coupled amplifier behaves like a low pass filter
Answer: voltage gain increases
Explaination:
when frequency increases, shunt reactance decreases. the voltage drop across shunt capacitance decreases and net voltage gain decrease. rc coupled amplifier acts as a low pass filter at high frequencies.
Q2 | In Miller’s theorem, what is the constant K?
  • total voltage gain
  • internal voltage gain
  • internal current gain
  • internal power gain
Answer: internal voltage gain
Explaination:
the constant k=v2/v1, which is the internal voltage gain of the network.
Q3 | When applying miller’s theorem to resistors, resistance R1 is for node 1 and R2 for node 2. If R1>R2, then for same circuit, then for capacitance for which the theorem is applied, which will be larger, C1 or C2?
  • c1
  • c2
  • both are equal
  • insufficient data
Answer: c1
Explaination:
given r1>r2
Q4 | Find net voltage gain, given hfe = 50 and hie = 1kΩ.
  • 27.68
  • -22
  • 30.55
  • -27.68
Answer: -27.68
Explaination:
apply millers theorem to resistance between input and output.
Q5 | Given that capacitance w.r.t the input node is 2pF and output node is 4pF, find capacitance between input and output node.
  • 0.67 pf
  • 1.34pf
  • 0.44pf
  • 2.2pf
Answer: 0.67 pf
Explaination:
c1=c(1-k), c2=c(1-k-1) c1=2pf
Q6 | 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.
  • 120
  • 12
  • 15 d) -12
Answer: 15 d) -12
Explaination:
av = -120
Q7 | 15Hz
  • 22.73 hz
  • 612 hz
  • 673hz
  • 317 hz
Answer: 612 hz
Explaination:
rc = 2kΩ, rl = 5kΩ, cc = 1μf, cb = 10μf, ce = 20μf, rs = 2 kΩ
Q8 | 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.
  • 50.8 hz
  • 55.9 hz
  • 60hz
  • 100hz
Answer: 55.9 hz
Explaination:
avm = 150 avl = 100
Q9 | What is the phase shift in RC coupled CE amplifier at lower 3dB frequency?
  • 180°
  • 225°
  • 270°
  • 100°
Answer: 225°
Explaination:
total phase shift = 180°+ tan- 1(fl/f)
Q10 | We cannot use h-parameter model in high frequency analysis because                          
  • they all can be ignored for high frequencies
  • junction capacitances are not included in it
  • junction capacitances have to be included in it
  • ac analysis is difficult for high frequency using it
Answer: junction capacitances are not included in it
Explaination:
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.
Q11 | 8 x 106.
  • 200
  • 100
  • 141.42
  • 440.2
Answer: 141.42
Explaination:
the current gain for the ce
Q12 | Which of the statement is incorrect?
  • at unity gain frequency the ce short circuit current gain becomes 1
  • unity gain frequency is the same as gain bandwidth product of bjt
  • gain of bjt decreases at higher frequencies due to junction capacitances
  • β- cut-off frequency is one where the ce short circuit current gain becomes β/2
Answer: at unity gain frequency the ce short circuit current gain becomes 1
Explaination:
gain bandwidth product for any mosfet is ft = gm/2π(cgs+cgd) thus gbp is approximately 5.9 mhz.
Q13 | The collector current will not reach the steady state value instantaneously because of                   
  • stray capacitances
  • resistances
  • input blocking capacitances
  • coupling capacitance
Answer: stray capacitances
Explaination:
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.
Q14 | 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?
  • 20ms
  • 50ms
  • 60ms
  • 70ms
Answer: 50ms
Explaination:
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
Q15 | The technique used to quickly switch off a transistor is by                   
  • reverse biasing its emitter to collector junction
  • reverse biasing its base to collector junction
  • reverse biasing its base to emitter junction
  • reverse biasing any junction
Answer: reverse biasing its base to emitter junction
Explaination:
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.
Q16 | Which of the following circuits helps in the applications of switching times?
  • c)
  • d)
Answer: d)
Explaination:
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.
Q17 | Which of the following helps in reducing the switching time of a transistor?
  • a resistor connected from base to ground
  • a resistor connected from emitter to ground
  • a capacitor connected from base to ground
  • a capacitor connected from emitter to ground
Answer: a resistor connected from base to ground
Explaination:
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.
Q18 | The time taken for a transistor to turn from saturation to cut off is                    
  • inversely proportional to charge carriers
  • directly proportional to charge carriers
  • charging time of the capacitor
  • discharging time of the capacitor
Answer: directly proportional to charge carriers
Explaination:
when sufficient charge carriers exist, the transistor goes into saturation.
Q19 | Switching speed of P+ junction depends on                    
  • mobility of minority carriers in p junction
  • life time of minority carriers in p junction
  • mobility of majority carriers in n junction
  • life time of minority carriers in n junction
Answer: life time of minority carriers in n junction
Explaination:
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.
Q20 | Fixed voltage regulators and adjustable regulators are often called as
  • series dissipative regulators
  • shunt dissipative regulators
  • stray dissipative regulators
  • all the mentioned
Answer: series dissipative regulators
Explaination:
series dissipative regulators simulate a variable resistance between the input voltage & the load and hence functions in a linear mode.
Q21 | Linear series regulators are suited for application with
  • high current
  • medium current
  • low current
  • none of the mentioned
Answer: medium current
Explaination:
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.
Q22 | A series switching regulators
  • improves the efficiency of regulators
  • improves the flexibility of switching
  • enhance the response of regulators
  • all of the mentioned
Answer: improves the efficiency of regulators
Explaination:
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.
Q23 | The switching regulators can operate in
  • step up
  • step down
  • polarity inverting
  • all the mentioned
Answer: all the mentioned
Explaination:
the switching regulators can operate in any one of the three modes depending on the way in which the components are connected.
Q24 | 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.
  • 1 and 3
  • 1,2,3 and 4
  • 2,3 and 4
  • 1,3 and 4
Answer: 1,2,3 and 4
Explaination:
a voltage source must satisfy the entire given requirement to be used in switching regulator.
Q25 | Which among the following act as a switch in switching regulator?
  • rectifiers
  • diode
  • transistors
  • relays
Answer: transistors
Explaination:
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.
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