Electrical machines miscellaneous
- A field excitation of 20 A in a certain alternator results in an armature current of 400 A in short circuit and a terminal voltage of 2000 V on open circuit. The magnitude of the internal voltage drop within the machine at a load current of 200 A is
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Synchronus impedance,
Zs = Vio = 2000 = 5 Ω ISC 400 
∴ Voltage drop = IZs
= 200 × 5 = 1000 V
Correct Option: D
Synchronus impedance,
Zs = Vio = 2000 = 5 Ω ISC 400
∴ Voltage drop = IZs
= 200 × 5 = 1000 V
- A separately excited dc machine is coupled to a 50 Hz, three-phase, 4-pole induction machine as shown in the figure. The dc machine is energized first and the machines rotate at 160 rpm. Subsequently the induction machine is also connected to a 50 Hz, three-phase source, the phase sequence being consistent with the direction of rotation. In steady state,
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Synchronus speed of induction machine,
= 120 × f = 120 × 50 = 1500 rpm P 4
Speed of d.c. machine, Nr = 1600 rpm = actual speed of IM∴ Slip = Ns - Nr = 1500 - 1600 = negative Ns 1500
Hence induction machine acts as induction generator and dc machine acts as motor.
Correct Option: C
Synchronus speed of induction machine,
= 120 × f = 120 × 50 = 1500 rpm P 4
Speed of d.c. machine, Nr = 1600 rpm = actual speed of IM∴ Slip = Ns - Nr = 1500 - 1600 = negative Ns 1500
Hence induction machine acts as induction generator and dc machine acts as motor.
- A single-phase transformer has a turns ratio of 1 : 2 and is connected to a purely resistive load as shown in the figure given below. The magnetizing current drawn is 1 A, and the secondary current is 1 A. If core losses and leakage reactances are neglected, the primary current is
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i 0 = 1A, i 2 = 1A,∴ i2' = n2 i2 = 2.1 = 2 A n1 1
i1 = √i0² + i2'²
i1 = √1² + 2² = √5 = 2.24 amp.
Correct Option: C
i 0 = 1A, i 2 = 1A,∴ i2' = n2 i2 = 2.1 = 2 A n1 1
i1 = √i0² + i2'²
i1 = √1² + 2² = √5 = 2.24 amp.
- The direct axis and quadrature axis reactances of a salient pole alternator are 1.2 p.u. and 1.0 p.u. respectively. The armature resistance is negligible. If this alternator is delivering rated KVA at upf and at rated voltage then its power angle is
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tan δ = Ia (xq cos φ + ra sin φ) Vt + Ia (xq sin φ - ra cos φ)
For, φ = 0° [using p.f.],
I a = Vt = 1 pu
Xd = 1.2 pu,
Xq = 1pu,
ra = 0
we get, tan δ = 1
⇒ δ = 45°
Correct Option: B
tan δ = Ia (xq cos φ + ra sin φ) Vt + Ia (xq sin φ - ra cos φ)
For, φ = 0° [using p.f.],
I a = Vt = 1 pu
Xd = 1.2 pu,
Xq = 1pu,
ra = 0
we get, tan δ = 1
⇒ δ = 45°
- A three-phase 440 V, 6 pole, 50Hz, squirrel cage induction motor is running at a slip of 5%. The speed of stator magnetic field with respect to rotor magnetic field and speed of rotor with respect to stator magnetic field are
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Ns = 120 × f = 120 × 50 = 1000 rpm P 6
Rotor speed = Ns – SNs = 950 r.p.m.
Stator magnetic field speed, Ns = 1000 r.p.m.
Rotor magnetic field speed, Ns = 1000 r.p.m.
Relative speed between stator and rotor magnetic fields is zero
Rotor speed with respect to stator magnetic field = 950 – 1000 = – 50 r.p.m.Correct Option: E
Ns = 120 × f = 120 × 50 = 1000 rpm P 6
Rotor speed = Ns – SNs = 950 r.p.m.
Stator magnetic field speed, Ns = 1000 r.p.m.
Rotor magnetic field speed, Ns = 1000 r.p.m.
Relative speed between stator and rotor magnetic fields is zero
Rotor speed with respect to stator magnetic field = 950 – 1000 = – 50 r.p.m.
