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A closed magnetic circuit of cast steel having a 6cm long path of cross-sectional area 1cm2 and a 2 cm path of cross-sectional area 0.5 cm2. A coil of 200 turns is wound around the 6 cm length of the circuit and a current of 0.4 A flows. Verify the flux density in the 2 cm path, if the relative permeability of the cast steel is 750.
A 10 - hp, 250-V shunt motor has an armature circuit resistance of 0.5 and a ?eld resistance of 200 . At no load, rated voltage, and 1200 r/min, the armature current is 3 A. At
Explain in detail the operation of 8255 in mode1 taking suitable example. In mode1, Ports A and B are programmed as input or output ports and Port C is used for handshaking.
Two three-phase, 6.6-kV, wye-connected synchronous generators, operating in parallel, supply a load of 3000 kWat 0.8 power factor lagging. The synchronous impedance per phase ofmac
Hexadecimal is of use in IT because (1) It is a compact system (e.g. only 3 digits represent the number 986) (2) As 16 are a power of 2 it turns out to be quite easy to conv
What is expanded memory system? EMS: The area at location C8000H-DFFFFFH is frequently open or free. Such area is used for the expanded memory system into a XT or PC system, or
Explain the Johnson Counters? The Johnson counters are a variation of standard ring counters with the inverted output of the last stage fed back to the input of the first stage
Using Bode plot calculate (a) Phase margin (b) Gain margin (c) Stability of closed loop system. The open transfer function of the system is t=30/(s+2)/(s+3)Using Bode plot calcula
A quantizer has 130 quantum levels that exactly span the extremes of a symmetrically ?uctuating message with step size δv = 0.04 V. Determine the following: (a) |f(t)|max.
Q. What do you mean by Load factor? The two components of electric service are demand and energy. Demand is the maximum level of real power which the electric utility must supp
Q. What is a positive clipper? Explain its action with the help of a circuit. A positive clipper is the circuit which is used to cut off the positive half cycle. The circuit wi
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