InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
Find current i. |
| Answer» The given circuit can be compared with Wheatstone's Bridge ∴Req = (10 + 10 + 10) || (10 + 10 + 10) = 30 || 30 = 15 Ω . | |
| 52. |
Consider a system with the transfer function . Its damping ratio will be 0.5 when the value of is : |
| Answer» ξ = 0.5, We take second term only . | |
| 53. |
For the system with given characteristic equation + 4 + 8 + 8 + 7 + 4 = 0 The number of roots on left hand side, right hand side and imaginary axis are respectively. |
| Answer» Apply Routh Hurwitz criteria. | |
| 54. |
A lossless 50 Ω transmission line is terminated 30 + 25 Ω, then voltage reflection coefficient and current reflection coefficient are |
| Answer» Voltage reflection coefficient Current reflection coefficient : ri = - rv = + 10.36 ∠111.296. | |
| 55. |
Find value of i |
| Answer» The given circuit can be compared to a Wheatstone's Bridge. This bridge is balanced. ∴ Req = (10 + 10) || (20 + 20) = 20 || 40 = 13.33 Ω ∴ . | |
| 56. |
Find the electric charge required on the earth and moon to balance their gravitational attraction if the charge on the earth is 12.5 times that on the moon.Mass of moon = 6.7 x 10 KMass of earth = 6 x 10 KDistance between moon and earth = 380 KUniversal gravitational constant = 6.7 x 10 N/K |
| Answer» Using law of gravitations, the force between two point masses is F = M1 = mass of moon = 6.7 x 1022 kg M2 = mass of earth = 6 x 1024 kg r = distance between masses = 380 km G = universal gravitational constant = 6.7 x 10- 11 Nm2/Kg2 Let Q1 and Q2 be the charge on moon and earth respectively. Now the gravitational force must be balanced by the force of repulsion. Q1Q2 = 4pε0M1M2G but Q2 = 12.5Q1 = 12.5 x 4pω0M1M2G ∴ Q2 = 6.11 x 1014 Q2 = 61.11 TC Q1 = 4.89 TC. | |
| 57. |
When () is real then the Fourier series coefficients are correlated as __________ . |
| Answer» a-k = a*k for all values of k. In fact, this equality holds whenever x[n] is real. | |
| 58. |
The electric field of 2.5 x 10 V/m can give 1 eV energy to condition electron, then in a copper block what will be the mean free path of electron? |
| Answer» The force on electron in eE. As the electron moves through distance d , the work done on it is eEd. This is equal to energy transferred to electron. ∴ eEd = 1 eV ∴ d = 4 x 10-7m. | |
| 59. |
25 persons are in a room. 15 of them play hockey, 17 of them play football and 10 of them play both hockey and football. Then the number of persons playing neither hockey nor football is: |
| Answer» Representing the given information in the Venn diagrams, we have Let the number of people who play only hockey = a The number of people who play only football = b Now, a = n(H) - 10 = 15 - 10 = 5 b = n(F) - 10 = 17 - 10 = 7 Clearly, a + b + 10 + n = 25 ⇒ n = 25 - 7 - 5 -10 ⇒ n = 3 ∴ The number of people who do not play neither Hockey nor Football is 3. | |
| 60. |
The transfer function of a system is . For a unit-step input to the system the approximate setting time for 2% criterion is : |
| Answer» ξωn = 1 for 2% = 4 sec. | |
| 61. |
The Nyquist sampling interval, for the signal sine (700) + sine (500) is |
| Answer» g(t) = sinc (700t) + sinc (500t) g(t) is band-limited with fM = 350 Hz. Hence the Nyquist rate will be 700 Hz. Then Nyquist interval is sec. | |
| 62. |
In the circuit shown, the device connected to Y5 can have address in the range |
| Answer» | |
| 63. |
The resonant frequency of the circuit shown in figure is, |
| Answer» Leq = L1 + L2 + 2M Leq = 1 + 2 + 2 x = 4 H ∴ . | |
| 64. |
A short grounded vertical antenna has a length L which is 0.05λ at frequency . If its radiation resistance at is R ohms, then its radiation resistance at a frequency 2 will be |
| Answer» Rr = 80p2 ∴ Rr ∝ f2. | |
| 65. |
A fair coil is tossed 10 times. What is the probability that ONLY the first two tosses will yield heads? |
| Answer» Use ncr, (p)r (q)n-r [for any two losses yield head]. But in present case it is required case it is only for first two tosses. Thus in this case 10 times. | |
| 66. |
The equation sin () = 10 has |
| Answer» Since value lies between - 1 and + 1. Therefore no real or complex solution exists. | |
| 67. |
Given If , then the value of K is |
| Answer» . | |
| 68. |
Consider the Schmitt trigger circuit shown below. A triangular wave which goes from - 12 V to 12 V is applied to the inverting input of the op-amp. Assume that the output of the op-amp moves from +15 V to -15 V. The voltage at the non-inverting switches between |
| Answer» V0 is switching between + 15 to - 150 Thus voltage at non-inverting input switches between , 7.5 to - 7.5 . | |
| 69. |
Determine the potential difference between the points A and B in the steady state |
| Answer» | |
| 70. |
The equation for loop currents I and I are __________ |
| Answer» k = 0.8 ∴ M = kL1L2 = 0.83 x 2 = 1.95 H Apply KVL to loop 1 200∠0° = (j5 + 5 + j2 - j3)I1 + 1.95I2 ∴ 200∠0° = (5 + j4)I1 + j1.95I2 Apply KVL to loop 2 0 = (5 + 3 + j3)I2 + j1.95I1 0 = (8 + j3)I2 + j1.95I1 0 = j1.95I1 + (8 + j3)I2 ∴ The loop equations in matrix form are | |
| 71. |
A power of 100 W is radiated from an isotropic radiator. The power radiated peer unit solid angle and the power density at a distance of 10 km from the antenna is respectively |
| Answer» . | |
| 72. |
It is desired to design a phase shift oscillator using an FET having a = 5000 μSiemen, = 40 Ω and feedback circuit value of R = 10 K. Then value of C for oscillator operation at 1 H and that of for A = 40 to ensure oscillator action are |
| Answer» For a FET oscillator, the oscillator frequency is ∴ C = 6.4 nF For FET Oscillator, |A| = gmRL = 5000 x 10-6 RL i.e., 40 = 5000 x 10-6 x RL ∴ = 8 KΩ Now, Using RL = 8 K and rd = 40 K (given) We find RD as follows: ∴ ∴ = 10 K RD = 10 K. | |
| 73. |
Find the value of K and velocity constant K so that the maximum overshoot in the unit step response is 0.2 and the peak time is 1 sec. |
| Answer» ζ2 = 0.259(1 - ζ2) = 0.259 - 0.259ζ2 1.259ζ2 = 0.259 ζ2 = 0.206 ζ = 0.45 ω2n = K ∴ ωn = K and 2ζωn = 1 + KKv ∴ ωn = K K = ω2n K = 12.37 1 + 12.37Kv = 2 x 0.45 x 3.52 Kv = 0.175 . | |
| 74. |
His rather casual remarks on politics __________ his lack of seriousness about the subject. |
| Answer» The key words in the statement are 'casual remarks' and 'lack of seriousness'. The blank should be filled with a word meaning 'showed' or 'revealed'. Hence, 'betrayed' is the correct answer. | |
| 75. |
A ramp input applied to a unity feedback system results in 5% steady state error. The type number and zero frequency gain of the system are respectively. |
| Answer» ess = 5% ⇒ .05 . | |
| 76. |
The resolution of a 4 bit counting ADC is 0.5 volts. For an analog input of 6.6 volts, the digital output of the ADC will be |
| Answer» Resolution = ⇒ = 0.5 2n = 14 = (1110) n = 4 bits. | |
| 77. |
The circuit given below is: |
| Answer» The given circuit is Band pass filter . | |
| 78. |
Rank of the matrix: |
| Answer» The given matrix is equivalent to , Thus rank is 1. | |
| 79. |
The ratio of the diffusion coefficient in a semiconductor has the units |
| Answer» . | |
| 80. |
Consider the probability density = where is a random variable whose allowable values range from = - ∞ to = + ∞. P(1 ≤ ≤ 2) |
| Answer» P(x) = ae-b|x|dx to get apply aebxdx + ae-bxdx, we get = . | |
| 81. |
The amplitude of a pair of composite sinusoidal signal() = () + () with () = sin (5) () = sin (5) is __________ . |
| Answer» Amplitude = . | |
| 82. |
A -channel JFET has I = 2 A, Gate to source voltage V = - 2 V and trans-conductance is 0.5 W then pinch-off voltage is __________ . |
| Answer» ∴ ∴ = - 8VP - 16 ∴ + 8VP + 16 = 0 ∴ (VP + 4)2 = 0 ∴ VP = -4 V. | |
| 83. |
In the circuit shown, in switch S is open for a long time and is closed at = 0. The current () for ≥ 0 is |
| Answer» i(f) = 0.5, i(i) = 0.75 i(t) = Vr + (ii - ij)e-1/p = 0.5 - 0.125e-1000t . | |
| 84. |
In a semiconductor material. The hole concentration is found to be 2 x 2.5 x 10 . If mobility of carriers is 0.13 / . Then find the current density if electric field intensity is 3.62 x 10 |
| Answer» Current density J = σE Where σ = conductivity Given -> μ = 0.13 m2/v-s = 0.13 x 104 cm2/V sec P = 2.25 x 1015/cm3 We have, ∴ ni = 1.5 x 1010 Also n.p. = ∴ n = /p = (1.6 x 10-19 x 0.13 x 104 x 2.25 x 1015) x = (0.468) (4.5 x 1015) σ = 2.106 x 1015 μ/cm J = σE ∴ Current density = 2.106 x 1015 x 3.620 x 10-19 = 7.6237 x 10-4 A/m2. | |
| 85. |
What is the propagation constant for air filled wave guide with dimensions = 1.59" and = 0.795" at 4.95 GHz? |
| Answer» Here, a = 1.59'' = 40.386 mm = 4.04 cm ∴ . | |
| 86. |
If () = then (0) is __________ (where ()⇔ () |
| Answer» The signal g(t) given above is a Gaussian pulse, and it satisfies the relation. Therefore, its fourier transform is same as the signal itself in frequency domain. i.e. G(f) = e-pf2 Hence G(f)|f = 0 = e-p(0) = 1. | |
| 87. |
The open loop transfer function of a unity feedback system is: The range of K for which the system is stable is |
| Answer» Apply R-H. | |
| 88. |
A discrete time linear shift-invariant system has an impulse response [] with [0] = 1, h[1] = - 1, [2] - 2, and zero otherwise. The system is given an input sequence [] with [0] - [2] - 1, and zero otherwise. The number of nonzero samples in the output sequence [], and the value of [2] are, respectively |
| Answer» Use convolution to get the result. | |
| 89. |
If we manage to __________ our natural resources, we would leave a better planet for our children. |
| Answer» The clue in this sentence is 'If we manage to __________ our natural resources' and 'better planet'. This implies that the blank should be filled by a word which means 'preserve' or 'keep for long time'. Therefore the word 'conserve' is the right answer. | |
| 90. |
(2) + (3) = (?) |
| Answer» (2)3 + (3)4 = 2 + 3 = 5 and (10)5 = 51 + 0 = 5 ∴ (2)3 + (3)4 = (10)5 . | |
| 91. |
A transmission line has a characteristic impedance of 50 Ω and a resistance of 0.1 Ω/m, if the line is distortion less, the attenuation constant (in Np/m) is |
| Answer» Distortion less a = RG and Z0 = RG → . | |
| 92. |
A certain 8 bit uniform quantization PCM system can accommodate a signal ranging from - 1 V to + V. The rms value of the signal is V . The signal to quantization noise ratio is: |
| Answer» (SNR)dB = 1.76 + 6.02n n = 8 Thus (SNR)dB = 1.76 + 6.02 x 8 = 49.92 dB. | |
| 93. |
A rectangular waveguide, in dominant TE mode, has dimensions 10 x 15 . The cut off frequency is |
| Answer» For TE10 mode | |
| 94. |
A radar receiver has a noise figure of 10 dB at 300 K having a bandwidth of 2.5 MHz. The minimum power it can receive is |
| Answer» Pm = kT0Δf(F - 1) = 93.15 x 10-15 W. | |
| 95. |
The circuit is shown in figure. The current transfer ratio , is |
| Answer» . | |
| 96. |
Let (A, B) = A ⊕ B; then the simplified form of the function (( ⊕ , )) is |
| Answer» f(A, B) = A ⊕ B f(x ⊕ y, z) = (x ⊕ y ⊕ z) = M But f(MW) is undefined. | |
| 97. |
A system has 12 poles or 2 zeros. Its high frequency asymptote in its magnitude plot has slope of |
| Answer» High frequency asymptotes = - 20(12) + 2(20) = - 200 dB/DECADE. | |
| 98. |
The maximum percent overshoot with maximum value of first overshoot as 3.5 V and steady state value as 2.575 V is |
| Answer» From the transient response characteristics, Maximum percentage overshoot (MPO) is given as MPO decides the relative stability of the system. | |
| 99. |
An XOR gate with 8 variable is as follows: A ⊕ B ⊕ C ⊕ D ⊕ E ⊕ F ⊕ G ⊕ HThe number of minterms in the Boolean expression is |
| Answer» . | |
| 100. |
The figure shown below is designed to act as a constant current source across the load resistor. The β of transistor is 50. What is the value of load current? |
| Answer» Voltage at base = 12 - 5 = 7 v Voltage at emitter = 7 + 0.7 = 7.7 v Current through R1 = 2 mA Thus current through load = = 1.96 mA. | |