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.
| 101. |
For a given magnetic material operating in linear mode B = 0.06 T. Assume permeability = 51, then magnetization is: |
| Answer» μ = λmH, λm = (μ - 1) = 51 - 1 = 50 μ = 50 x H 46834. | |
| 102. |
(1 2 1) = (1 0 0) Then is equal to |
| Answer» x2 = 2x + 1 = 16 x2 + 2x - 15 = 0, (x - 3)(x + 5) = 0, x = 3, - 5 Thus x = 3. | |
| 103. |
For the network shown below, the Thevenin voltage and R is equal to __________ |
| Answer» Since there is no independent source in the network; VTH = 0. To find RTH apply 1 A current source across A and B. Apply KCL at node Vx as Solving Vx = 58.82 V ∴ . | |
| 104. |
Impedance function for the network shown below is: |
| Answer» | |
| 105. |
The function = A ⊕ B ⊕ ⊕ D ⊕ E ⊕ C ⊕ ⊕ E ⊕ F ⊕ G ⊕ F ⊕ G ⊕ C ⊕ E ⊕ A ⊕ D ⊕ ⊕ E ⊕ ⊕ ⊕ B ⊕ ⊕ B ⊕ D can be written as |
| Answer» A A B B C D D E E F G 2 1 3 3 2 3 1 4 1 2 2 0 A B B 0 D D 0 E 0 0 f = A ⊕ B ⊕ B ⊕ D ⊕ D ⊕ E = A ⊕ 1 ⊕ 1 ⊕ E = A ⊕ 0 ⊕ E = A ⊕ E. | |
| 106. |
Find the maximum power that can be delivered to a load |
| Answer» PL = I2R = x R •••(i) where zS = 100 + j50 zL = 100 - j50 Putting the values in eq. (i) PL =X l00 = 0.25 watts. | |
| 107. |
For a integrator circuit, R = 100 KΩ, C = 10 μF. The input is a step voltage as shown below. Its output will be |
| Answer» | |
| 108. |
The Causal system given below is __________ () = 6 + - |
| Answer» Minimum phase H(z) = 6 + z-1 - z-1 6z2 + z - 1 = 0 ∴ We have zeros at z = 1/3 and z = - 1/2 ∴ All the zeroes lies inside the unit circle, it is a min. phase system. | |
| 109. |
Given potential sin Θ cos Φ, find flux density D at (2, 2, 0) |
| Answer» Now E = -∇V E = ∇V At r = 2, θ = p/2; Φ = 0 we have ∴ D = 22.12 PC/m2. | |
| 110. |
The distance between target and radar is reduced to half. Power received increases by |
| Answer» . | |
| 111. |
If the memory chip size is 256 x 1 bits, then the number of chips required to make up 1 kB (1024) bytes of memory is |
| Answer» . | |
| 112. |
A coil has a designed for high Q performance at a rated voltage and a specified frequency. If the frequency of operation is doubled, and the coil is operated at the same rated voltage, then the Q factor and the active power P consumed by the coil will be affected as follows. |
| Answer» ω2L = 2ω1L, R remains constant ∴ Q2 = = 2Q1 i.e. Q is doubled for a high a coil ωL >> R ∴ i.e. P decreases 4 times. | |
| 113. |
Find the number of loop equation and number of possible trees for the given graph. |
| Answer» In any graph, Number of loop equations L = B - N + 1 where B = Number of branches, N = Number of nodes. Given B = 14, N = 8 ∴ L = B - N + 1 = 7 The total number of trees possible T = NN - 2 = 88 - 2 = 262144. | |
| 114. |
The number of open right half plane poles of is |
| Answer» Use Routh criteria. | |
| 115. |
If the sequence converges, then the series converges absolutely is |
| Answer» This is the Root test. Since the signal magnitude is raised to the power of 1/n. | |
| 116. |
Open loop transfer function of a system isThe phase cross over frequency ω is |
| Answer» GH = - tan- 1 ω - tan- 1 2ω - tan- 1 3ω = - p tan- 1 ω + tan- 1 2ω = p - tan- 1 3ω tan- l = p - tan- 1 3ω = - 3ω ω = 1 rad/sec. | |
| 117. |
A λ/4 long high frequency transmission line is terminated into one impedance Z. If Z be the characteristic impedance of the line, then input impedance Z is |
| Answer» , here and ∴ [Here ZL = ZR]. | |
| 118. |
The function shown in figure, can represent a probability density function for A __________ |
| Answer» The condition for the probability density function d(x) isand is always non negative or the area under the curve f(x)is unity. => A x 6 +x 2 x 3A = 1 or 9A = 1 or A =. | |
| 119. |
A bipolar differential amplifier uses a transistor having β = 200, I = 100 A, |A| = 500 and CMRR = 80 dB. The value of R and R will be respectively __________ . |
| Answer» 80 = 20 log CMRR ∴ CMRR = 10000 But CMRR = 2gmRE ∴ = 1.25 MΩ Adm = - gmRC = 1.25 KΩ. | |
| 120. |
When phase-lag compensation is used in a system, then gain cross over frequency, bandwidth and undamped frequency are respectively |
| Answer» Phase lag is delay between a correcting signal of control system and response to it. | |
| 121. |
The equivalent form of the logical expression B + C + ABC + AB + A is |
| Answer» This can be solve by using Boolean identity but using k map it will be more easy (A + B + C) (A + B + C) (A + B + C). | |
| 122. |
The gradient of any scalar field always yields __________ . |
| Answer» Dot product is conservative. Gradient is nothing but dot product. | |
| 123. |
In a flash type ADC, employing 15 comparators resolution with 10 volts reference; is expected to be : |
| Answer» Since number of comparators = 2n - 1 = 15; Then number of bits are '4' Resolution | |
| 124. |
Radiation resistance of an antenna is 54 Ω and loss resistance is 6 Ω. If antenna has power gain of 10, then directivity is: |
| Answer» Efficiency of antenna Power gain = 10 Thus directivity . | |
| 125. |
If the power spectral density of stationary random process is a sine-squared function of frequency, the shape of its autocorrelation is |
| Answer» Since autocorrelation function and power spectral density bears a Fourier transform relation, then since required in frequency domain will five rectangular convolutions in time domain thus it is triangular function. | |
| 126. |
The inverse of given Laplace transform is |
| Answer» s = x2 - ex x(t) = cos t. | |
| 127. |
Consider the amplitude modulated (AM) signal cos ω + 2 cos ω cos ω For demodulating the signal using envelope detector, the minimum value of A should be |
| Answer» AC cos ωct + 2 cos ωmt cos ωct AC cosωct for envelope detection μ < 1 ⇒ < 1 ⇒ Ac should be at least-2. | |
| 128. |
In a 4 bit weighted resistor D/A converter, the resistor value corresponding to LSB is 32 kΩ. The resistor value corresponding to MSB will be |
| Answer» 2n - 1 R = 32 K&Omeg; n = 4 ⇒ 8R = 32 R = 4 KΩ. | |
| 129. |
Find in the circuit below :(A, B, C, D) = Σ(6, 7, 13, 14);(A, B, C, D) = Σ(3, 6, 7);(A, B, C, D) = Σ(5, 6, 7, 14, 15) |
| Answer» f1(A, B, C, D) = ∑(6, 7, 13, 14) f2(A, B, C, D) = ∑(3, 6, 7) f1 ⊕ f2 = ∑(3, 13, 14) f3 x (f1 ⊕ f2) = ∑(14) = y X = f1 x y ⇒ ∑14. | |
| 130. |
What are the values of E and E displayed on the oscilloscope, when a 1 kV - carries is modulated to 50%? |
| Answer» Emax - Emin = 0.5 x 1 kV = 0.5 kV 2Emax = 1.5 kVi, Emax = 0.75 kV Emin = 0.25 kV. | |
| 131. |
As the temperature increases the mobility of electrons __________ . |
| Answer» Mobility of electrons m = 2.5 for electrons and 2.7 for holes in Si. | |
| 132. |
A conductor in plane and having length 1 m is moving with a velocity V = (2 + 3 + ) m/sec. A magnetic induction field B = ( + 2) Wb/m is applied to the conductor. The potential difference between the ends of the conductor is |
| Answer» = - (V x B) = - (2i + 3j + k) x (i + 2j) ∴ || = (E2x + E2y + E2z) = 6 v/m ∴ Potential Difference across 1 m length = 6. | |
| 133. |
A 1000 H carrier wave modulated 40% at 4000 H is applied to a resonant circuit tuned to a carrier frequency and having = 140. What is the degree of modulation after passing the wave through this circuit? |
| Answer» Resulting depth of modulation is given by : when δ = fc = 1000 x 103 Hz fm = 4 x 103 Hz δ = m0 = = = 0.27 . | |
| 134. |
System is at rest for < 0. V = sin 2. Determine () for greater than zero. |
| Answer» KCL at node V ∴ Natural response from s + 2, is Ae-2t For Vs = sin 2t, s = 2j Steady state response Since the system is at rest for t < 0 Now substituting t = 0, we get | |
| 135. |
A 300 MHz plane EM wave is propagating in free space. The wave is incident normally on an infinite copper slap. The antenuation constant for the wave is (σ = 5.8 x 10 mho/ m). The skin depth is |
| Answer» . | |
| 136. |
Following figure represents |
| Answer» The output C is low only when A and B are both 1. | |
| 137. |
The practical resonant circuit is shown in above figure. Find the expression for the resonant frequency ω |
| Answer» Rationalising, Equating imaginary parts of numerator to 0, . | |
| 138. |
The common emitter model is shown below: The h-parameter of this model are: |
| Answer» h parameter V1 = (3 + 5)I1 + V2 . | |
| 139. |
Four signals = cos (ω), () = cos (ω), = 2 cos (ω) and = cos (4 ω) are multiplexed by Time Division Multiplexing system. The commutator speed in rps is |
| Answer» Apply Nyquist criteria. Maximum angular speed is 4ω0 Thus commutator speed must be 2 x 4ω2 = 8ω0 . | |
| 140. |
The feedback control system shown in figure is stable for what values of K and T? |
| Answer» Apply Routh Hurwitz criteria. | |
| 141. |
Find the value of if each resistance is of 3 Ω and each battery is of 6 V. |
| Answer» Using Mesh analysis Loop 1 : 6 = 9i1 - 3i4 ...(i) Loop 2: 6 + 6 = - 6i2 i2 = - 2A ...(ii) Loop 3: 6 = 9i3 - 3i4 ...(iii) Loop 4: 6 = 9i4 - 3i3 - 3i1 ...(iv) Solving the above equations simultaneously i1 = i3 = 1.1428 A i4 = 1.432 A ∴ ix = i4 - i3 = 0.2892 A. | |
| 142. |
If then for this to be true () is __________ . |
| Answer» Consider x(t) = ej2pft y(t) = h(t) * x (t) ⇒ y(t) = h(t). ej2pf(t - t) . dt = ei2pfth(t) . e- j2pft . dt = ej2pft x H(f) = x(t) x H(f) ⇒ . | |
| 143. |
The dual transform of the given network as |
| Answer» Dual of R → G | |
| 144. |
The transfer function Y()/R() of the system shown is |
| Answer» Construct a signal flow graph . | |
| 145. |
For = - 1, find the value of V |
| Answer» V1 = i1R1 + i2R2 i1 + ai1 = i2 V2 = i2R2 Putting a = - 1 = 0 V. | |
| 146. |
A unity-feedback control system has the open loop transfer function . If the input system is a unit ramp, the steady state error will be |
| Answer» . | |
| 147. |
For a given voltage, four heating coils will produce maximum heat when connected |
| Answer» . | |
| 148. |
Denominator polynomial of a transfer function of certain network is: + + 2 + 24 Then the network is: |
| Answer» Routh array There is negative number present in first column. Thus network is unstable. | |
| 149. |
A TV picture is to be transmitted over a channel of 6 MH bandwidth at a 35 dB S/N ratio. The capacity of the channel is |
| Answer» W = B = Bandwidth C = 6 x 106 x 12 = 72 Mbps. | |
| 150. |
If then the sequence is |
| Answer» Consider Number of samples per period = 10p This is not a rational number. Hence signal is non periodic. | |