2 Marks
In the circuit shown, Vc is 0 Vatt = 0 s. Fort >0,the capacitor current ic(t), where t is in second is given by
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Solution
2 Marks
Consider the TF of the following network as \(\frac{V_{0}(s)}{v_{i}(s)}=\frac{1}{3+sCR}\).Then the value of RL is
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Solution
2 Marks
In the figure shown below, assume that all the capacitors are initially uncharged. If Vi(t) = 6u(t), then Vo(t) at steady state is given by
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Solution
2 Marks
Consider the following statements relating to a circular disc rotating in a transverse magnetic field B Wb/m2 as shown in figure.The generated emf across outer rim A and at centre 0 is proportional to
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Solution
The generated emf across outerrim A and at centre O is proportional to — Angular velocity
— Flux density
— Square of the radius of disc.
2 Marks
A 200 MVA, 4 pole, 50 Hz alternator has a moment of inertia of 60 × 103 kqm2, The angular momentum of the rotor in (MJ s/elec degree) is ________.
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Solution
2 Marks
A system is shown with input signal x(t) and output signal y(t) ×x(t) has Fourier transform X (jω) as shown
The spectrum of y (jω) will be
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Solution
The spectrum for the given input signal.
2 Marks
A 500 V,25 HP ,DC shunt motor takes a current of 2.4 Awhile running lightly loaded. The field and armature resistances are 650 ohm and 0.57 ohm, respectively.Assume a brush drop of 2 V.
The constant losses of the machine are
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Solution
2 Marks
A generator feeds power to an infinite bus through a double circuit transmission line. A3-phase fault occurs at the middle point of the lines. The infinite bus voltage is 1 pu, the transient interval voltage of generator is 1.2 pu.The 100 MVA generator has an inertia constant of 5mJ/MVA and it was 1.0 PU power prior to the fault with rotor power angle of 45°. The system frequency is 50 Hz.The initial accelerating power (in pu) will be
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Solution
2 Marks
Identify the correct phase plot (approx) for H(s)\(=\frac{s+1000}{(s+10)^{2}}\)
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Solution
2 Marks
An induction motor has ratings 3 phase 50 Hz 6-pole shaft output 10 kW at 930 rpm Maximum torque occurs at 800 rpm.Starting torque (neglecting friction and windage loss) will be
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Solution