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This section includes 252 Mcqs, each offering curated multiple-choice questions to sharpen your Electrical Engineering knowledge and support exam preparation. Choose a topic below to get started.
101. |
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» D. reverse biasing any junction | |
102. |
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» B. resistances | |
103. |
What is the phase shift in RC coupled CE amplifier at lower 3dB frequency? |
A. | 180° |
B. | 225° |
C. | 270° |
D. | 100° |
Answer» C. 270° | |
104. |
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» C. 60hz | |
105. |
15Hz |
A. | 22.73 hz |
B. | 612 hz |
C. | 673hz |
D. | 317 hz |
Answer» C. 673hz | |
106. |
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» D. | |
107. |
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» B. 1.34pf | |
108. |
Find net voltage gain, given hfe = 50 and hie = 1kΩ. |
A. | 27.68 |
B. | -22 |
C. | 30.55 |
D. | -27.68 |
Answer» E. | |
109. |
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? |
A. | c1 |
B. | c2 |
C. | both are equal |
D. | insufficient data |
Answer» B. c2 | |
110. |
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 |
Answer» B. unity gain frequency is the same as gain | |
111. |
In Miller’s theorem, what is the constant K? |
A. | total voltage gain |
B. | internal voltage gain |
C. | internal current gain |
D. | internal power gain |
Answer» C. internal current gain | |
112. |
In an RC coupled CE amplifier, when the input frequency increases, which of these are incorrect? |
A. | reactance csh decreases |
B. | voltage gain increases |
C. | voltage gain decreases due to shunt capacitance |
D. | an rc coupled amplifier behaves like a low pass filter |
Answer» C. voltage gain decreases due to shunt capacitance | |
113. |
Gain bandwidth frequency is GBP= 3000 Mhz. The cut-off frequency is f=10Mhz. What is the CE short circuit current gain at the β cutoff frequency? |
A. | 212 |
B. | 220 |
C. | 300 |
D. | 200 |
Answer» B. 220 | |
114. |
8 x 106. |
A. | 200 |
B. | 100 |
C. | 141.42 |
D. | 440.2 |
Answer» D. 440.2 | |
115. |
Given that transition capacitance is 5 pico F and diffusion capacitance is 80 pico F, and base emitter dynamic resistance is 1500 Ω, find the β cut-off frequency. |
A. | 7.8 x 106 rad/s |
B. | 8.0 x 106 rad/s c) 49.2 x 106 rad/s d) 22.7 x 106 rad/s |
Answer» B. 8.0 x 106 rad/s c) 49.2 x 106 rad/s d) 22.7 x 106 rad/s | |
116. |
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» C. junction capacitances have to be included in it | |
117. |
If a C.E. stage has a load Rl and transconductance gm, what is the factor by which the capacitance between the base and the collector at the output side gets multiplied? |
A. | 1 + 1/gmrl |
B. | 1 – 1/gmrl |
C. | 1 + 2/gmrl |
D. | 1 – 2/gmrl |
Answer» B. 1 – 1/gmrl | |
118. |
If a C.E. stage has a load Rl and transconductance gm, what is the factor by which the capacitance between the base and the collector at the input side gets multiplied? |
A. | 1 + gmrl |
B. | 1 – gmrl |
C. | 1 + 2*gmrl |
D. | 1 – 2*gmrl |
Answer» B. 1 – gmrl | |
119. |
Which effect plays a critical role in producing changes in the frequency response of the B.J.T.? |
A. | thevenin’s effect |
B. | miller effect |
C. | tellegen’s effect |
D. | norton’s effect |
Answer» B. miller effect | |
120. |
If the total capacitance between the base and the emitter increases by a factor of 2, the transit frequency |
A. | reduces by 2 |
B. | increases by 2 |
C. | reduces by 4 |
D. | increases by 4 |
Answer» B. increases by 2 | |
121. |
If the transconductance of the B.J.T increases, the transit frequency |
A. | increases |
B. | decreases |
C. | doesn’t get affected |
D. | doubles |
Answer» B. decreases | |
122. |
Which parasitic capacitors don’t affect the frequency response of the C.C. stage of the B.J.T.? |
A. | ccs |
B. | ccs and cb |
C. | cb |
D. | ccs and cµ |
Answer» B. ccs and cb | |
123. |
Which parasitic capacitors don’t affect the frequency response of the C.B. stage of the B.J.T.? |
A. | none of the parasitic capacitances |
B. | all the parasitic capacitances |
C. | some of the coupling capacitors |
D. | ccs and cb |
Answer» C. some of the coupling capacitors | |
124. |
Which parasitic capacitors do not affect the frequency response of the C.E. stage, of the B.J.T.? |
A. | cje and cb |
B. | ccs and cµ |
C. | cb and cµ |
D. | no parasitic capacitor gets deactivated |
Answer» E. | |
125. |
3 BJT FREQUENCY RESPONSE |
A. | cje and cb |
B. | ccs and cµ |
C. | cb |
D. | ccs and cb |
Answer» C. cb | |
126. |
During high frequency applications of a B.J.T, which parasitic capacitors arise between the collector and the base? |
A. | cje and cb |
B. | ccs |
C. | cπ |
D. | cµ |
Answer» E. | |
127. |
During high frequency applications of a B.J.T., which parasitic capacitors arise between the collector and the emitter? |
A. | no capacitor arises |
B. | ccs |
C. | cb |
D. | ccs and cb |
Answer» B. ccs | |
128. |
During high frequency applications of a B.J.T., which parasitic capacitors arise between the base and the emitter? |
A. | cje and cb |
B. | ccs |
C. | cb |
D. | ccs and cb |
Answer» B. ccs | |
129. |
Under what condition can the circuit shown be called a compensated attenuator. |
A. | c1r1 = c2r2 |
B. | c1r2 = c2r1 |
C. | c1c2 = r1r2 |
D. | r1 = 0 |
Answer» B. c1r2 = c2r1 | |
130. |
General representation of the frequency response curve is called |
A. | bode plot |
B. | miller plot |
C. | thevenin plot |
D. | bandwidth plot |
Answer» B. miller plot | |
131. |
Which of the following is not a classification of amplifiers on the basis of their frequency response? |
A. | capacitively coupled amplifier |
B. | direct coupled amplifier |
C. | bandpass amplifier |
D. | none of the mentioned |
Answer» E. | |
132. |
STC networks can be classified into two categories: low-pass (LP) and high-pass (HP). Then which of the following is true? |
A. | hp network passes dc and low frequencies and attenuate high frequency and opposite for lp network |
B. | lp network passes dc and low frequencies and attenuate high frequency and opposite for hp network |
C. | hp network passes dc and high frequencies and attenuate low frequency and opposite for lp network |
D. | lp network passes low frequencies only and attenuate high frequency and opposite for hp network |
Answer» C. hp network passes dc and high frequencies and attenuate low frequency and opposite for lp network | |
133. |
Consider a voltage amplifier having a frequency response of the low-pass STC type with a dc gain of 60 dB and a 3-dB frequency of 1000 Hz. Then the gain db at |
A. | f = 10 hz is 55 db |
B. | f = 10 khz is 45 db |
C. | f = 100 khz is 25 db |
D. | f = 1mhz is 0 db |
Answer» E. | |
134. |
The signal whose waveform is not effected by a linear circuit is |
A. | triangular waveform signal |
B. | rectangular waveform signal |
C. | sine/cosine wave signal |
D. | sawtooth waveform signal |
Answer» D. sawtooth waveform signal | |
135. |
In a small signal differential gain vs input CM level graph, the gain decreases after V2 due to: |
A. | as the input voltage increases, the output will be clipped |
B. | when the input voltage to the transistors are high, the transistor enters saturation region and increases the current, which inturn decreases the output voltage = vdd – rd.iss |
C. | when common mode voltage is greater than or equal to v2, the input transistors enter triode region, the gain begins to fall |
D. | increasing the input voltage beyond v2 causes the gate oxide to conduct and the gain is reduced |
Answer» D. increasing the input voltage beyond v2 causes the gate oxide to conduct and the gain is reduced | |
136. |
In Common Mode Differential Amplifier, the outputs Vout1 and Vout2 are related as: |
A. | vout2 is in out of phase with vout1 with same amplitude |
B. | vout2 and vout1 have same amplitude but the phase difference is 90 degrees |
C. | vout1 and vout2 have same amplitude and are in phase with each other and their respective inputs |
D. | vout1 and vout2 have same amplitude and are in phase with each other but out of phase with their respective inputs |
Answer» E. | |
137. |
The inputs to the differential amplifier are applied at |
A. | at x and y |
B. | at the gates of m1 and m2 |
C. | all of the mentioned |
D. | none of the mentioned |
Answer» C. all of the mentioned | |
138. |
If source resistance in an amplifier circuit is zero, then voltage gain (output to input voltage ratio) and source voltage gain (output to source voltage ratio) are the same. |
A. | true |
B. | false |
Answer» B. false | |
139. |
The Differential output of the difference amplifier is the amplification of |
A. | difference between the voltages of input signals |
B. | difference between the output of the each transistor |
C. | difference between the supply and the output of the each transistor |
D. | all of the mentioned |
Answer» B. difference between the output of the each transistor | |
140. |
Consider the following circuit, where source current = 10mA, hfe = 50, hie = 1100Ω, then for the transistor circuit, find output resistance RO and input resistance RI. |
A. | ro = 0, ri = 21Ω |
B. | ro = ∞, ri = 0Ω |
C. | ro = ∞, ri = 21Ω |
D. | ro = 10, ri = 21Ω |
Answer» D. ro = 10, ri = 21Ω | |
141. |
Given that IB = 5mA and hfe = 55, find load current. |
A. | 28ma |
B. | 280ma |
C. | 2.5a |
D. | 2a |
Answer» C. 2.5a | |
142. |
For the given circuit, input resistance RI = 20Ω, hfe = 50. Output resistance = ∞. Find the new values of input and output resistance, if a base resistance of 2kΩ is added to the circuit. |
A. | ri = 20Ω, ro = ∞ |
B. | ri = 20Ω, ro = 2kΩ |
C. | ri = 59Ω, ro = ∞ |
D. | ri = 59Ω, ro = 2kΩ |
Answer» D. ri = 59Ω, ro = 2kΩ | |
143. |
Given hfe = 60, hie=1000Ω, hoe = 20μ Ω–, hre = 2 * 10-4. Find the current gain of the BJT, correct up to two decimal points. |
A. | – 58.44 |
B. | -59.21 |
C. | – 60.10 |
D. | – 60.00 |
Answer» B. -59.21 | |
144. |
Consider the circuit. Given hfe = 50, hie = 1200Ω. Find voltage gain. |
A. | – 278 |
B. | -277.9 |
C. | – 300 |
D. | – 280 |
Answer» B. -277.9 | |
145. |
Given yfs = 3.6mS and yos = 0.02mS, determine r0? |
A. | 100kohm |
B. | 50mohm |
C. | 50kohm |
D. | 20kohm |
Answer» D. 20kohm | |
146. |
In a small signal equivalent model of an FET, What does gm VGS stand for? |
A. | a pure resistor |
B. | voltage controlled current source |
C. | current controlled current source |
D. | voltage controlled voltage source |
Answer» C. current controlled current source | |
147. |
A FET has IDSS=4ID and gm0 = 10mS then gm = |
A. | 10ms |
B. | 20ms |
C. | 5ms |
D. | 14ms |
Answer» D. 14ms | |
148. |
Find the maximum value of gm for FET with IDSS=10mA, Vp=-2V, VGS=5V? |
A. | 10ms |
B. | 20ms |
C. | 1ms |
Answer» B. 20ms | |
149. |
002. What is the value ofgm? |
A. | 1 |
B. | 2 |
C. | 0.002 |
Answer» D. | |
150. |
Determine the value of output impedance for JFET, if the value of gm =1mS? |
A. | 1kohm |
B. | 0 |
C. | 100kohm |
D. | 5kohm |
Answer» B. 0 | |