Q. The Boolean function 𝐹( 𝑋, 𝑌) realized by the given circuit is (A) 𝑋̅ 𝑌 + 𝑋 𝑌̅                 (B)  ̅𝑋 𝑌̅ + 𝑋 𝑌              (C)  𝑋 + 𝑌                   (D)  𝑋̅  ∙ 𝑌̅ Ans: X̅ Y + X Y̅

Q. A 2-bit synchronous counter using two J-K flip flops is shown. The expressions for the inputs to the J-K flip flops are also shown in the figure. The output sequence of the counter starting from Q1Q2 = 00 is (A) 00 → 11 → 10 → 01 → 00 … (B) 00 → 01

Q. The product of sum expression of a Boolean function 𝐹(𝐴, 𝐵, 𝐶) of three variables is given by 𝐹(𝐴, 𝐵, 𝐶) = ( 𝐴 + 𝐵 + 𝐶̅ ) ∙ (𝐴 + 𝐵̅ + 𝐶̅) ∙ ( 𝐴̅ + 𝐵 + 𝐶 ) ∙ ( 𝐴̅ + 𝐵̅ + 𝐶̅ ) The canonical sum

Q. The circuit given uses ideal opamps. The current I (in mA) drawn from the source vs is (up to two decimal places) _  . Ans: 0.98 – 1

Q. An opamp that is powered from a ± 5V supply is used to build a non-inverting amplifier having a gain of 15. The slew rate of the opamp is 0.5*106 V/s. For a sinusoidal input with amplitude of 0.3 V, the maximum frequency (in kHz) up to which it can be operated without any

Q. Unit step response of a linear time invariant (LTI) system is given by 𝑦(𝑡) = (1 − 𝑒−2𝑡)𝑢(𝑡). Assuming zero initial condition, the transfer function of the system is Ans: Option D

Q. The Fourier transform of a signal 𝑥(𝑡), denoted by 𝑋(𝑗w), is shown in the figure. Let 𝑦(𝑡) = 𝑥(𝑡) + 𝑒𝑗𝑡𝑥(𝑡). The value of Fourier transform of 𝑦(𝑡) evaluated at the angular frequency w= 0.5 rad/s is (A) 0.5                        (B) 1                           (C) 1.5                       (D) 2.5 Ans: 1.5

Q. In the given circuit, the mesh currents I1, I2 and I3 are (A) I1 = 1 A, I2 = 2 A and I3 = 3 A                 (B) I1 = 2 A, I2 = 3 A and I3 = 4 A (C) I1 = 3 A, I2 = 4 A and I3 = 5 A                

Q. In the figure, an RLC load is supplied by a 230 V, 50 Hz single phase source. The magnitude of the reactive power (in VAr) supplied by the source is Ans: 66-68

Q. In the given circuit, superposition is applied. When V2 is set to 0 V, the current I2 is -6 A. When V1 is set to 0 V, the current I1 is +6 A. Current I3 (in A) when both sources are applied will be (up to two decimal places) . Ans: 0.95-1.05