A squirrel cage induction motor is to be started using a star-delta starter. Compared to direct-on-line (DOL) starting, the starting torque with star-delta starting is:
Answer: A
In star-delta starting, the motor is first connected in star, reducing the voltage across each winding to V/3 of the line voltage. Since torque is proportional to the square of the voltage applied to each winding:
Tstar=(31)2TDOL=31TDOL
Similarly, the starting current drawn from the supply is also reduced to 31 of the DOL starting current. Hence Tstar=31TDOL.
Q.662Medium
In a chopper-controlled DC drive, a DC motor is supplied from a 200 V source through a chopper with a duty cycle of 0.6. Assuming continuous current conduction, the average output voltage applied to the motor armature is:
Answer: A
For a step-down (buck) chopper, the average output voltage is:
Vo=δ×Vs
where δ is the duty cycle and Vs is the source voltage.
Vo=0.6×200=120V
Q.663Medium
The torque-speed characteristic of a fan or centrifugal pump type load follows the relationship:
Answer: A
For fans, blowers, and centrifugal pumps, the torque required varies as the square of the speed:
T∝N2
Consequently, the power consumed varies as the cube of speed: P∝N3. This is in contrast to constant torque loads (e.g., conveyors, hoists) where T=constant. This characteristic makes VFDs particularly effective for fan/pump drives, offering significant energy savings at reduced speeds.
Q.664Medium
In vector (field-oriented) control of an induction motor, the stator current is resolved into two decoupled components. These components are responsible for:
Answer: A
In Field-Oriented Control (FOC), the stator current vector is decomposed into two orthogonal components in a rotating reference frame aligned with the rotor flux:
This decoupling allows the induction motor to be controlled like a separately excited DC motor, achieving fast dynamic response. The torque is given by Te∝ψr⋅iqs, where ψr is the rotor flux controlled by ids.
Q.665Medium
A point charge Q=4μC is located at the origin. The electric flux through a spherical surface of radius r=0.5m centered at the origin is:
Answer: C
By Gauss's law, the total electric flux through any closed surface enclosing charge Q is
ΦE=∮E⋅dS=ε0Qenc.
Here Qenc=4μC=4×10−6C, so
ΦE=8.854×10−124×10−6≈4.52×105V⋅m.
The radius of the surface does not affect the total flux; only the enclosed charge matters.
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Q.666Medium
In a uniform plane electromagnetic wave propagating in free space, if the peak electric field intensity is E0=120πV/m, the peak magnetic field intensity H0 is:
Answer: A
The intrinsic impedance of free space is
η0=ε0μ0≈120πΩ.
The relationship between peak fields is
H0=η0E0=120π120π=1A/m.
Q.667Medium
The boundary condition for the normal component of the magnetic flux density B at the interface between two magnetic media (medium 1 and medium 2) with no surface current is:
Answer: B
From ∇⋅B=0 (Gauss's law for magnetism), applying the divergence theorem to a pillbox surface at the interface gives
B1n−B2n=0⟹B1n=B2n.
The normal component of B is always continuous across any interface. Note that Hn is not necessarily continuous when μ1=μ2.
Q.668Medium
The attenuation constant α and phase constant β of a lossy transmission line are related to line parameters R,L,G,C per unit length by the propagation constant γ=α+jβ. For a distortionless line, the condition that must be satisfied is:
Answer: C
A distortionless (Heaviside) line requires that the attenuation α is frequency-independent and the phase velocity vp=ω/β is constant (no dispersion). This is achieved when
LR=CG,
which rearranges to
RC=LG.
Under this condition, α=RG and β=ωLC, giving a flat frequency response.
Q.669Medium
The skin depth δ in a good conductor at frequency f is given by δ=ωμσ2. If the skin depth of copper at 1MHz is 0.066mm, what is the skin depth at 100MHz?
Answer: B
Skin depth varies as
δ∝f1.
Therefore,
δ1δ2=f2f1=100MHz1MHz=101.
So
δ2=100.066=0.0066mm.
As frequency increases, skin depth decreases, confining current to a thinner surface layer.
Q.670Medium
Which of the following is the correct differential form of Faraday's law of electromagnetic induction in a medium?
Answer: B
Faraday's law in differential (point) form is one of Maxwell's four equations:
∇×E=−∂t∂B.
The negative sign is a consequence of Lenz's law — the induced electric field opposes the change in magnetic flux. Option A is missing the negative sign; option C incorrectly uses the divergence operator; option D is Ampere's law with a wrong sign.
Q.671Medium
Two infinitely long parallel wires carry currents I1=10A and I2=10A in the same direction and are separated by a distance d=0.1m. The force per unit length between the wires is:
Answer: C
The force per unit length between two parallel current-carrying conductors is
Currents in the same direction attract each other (by the right-hand rule, each wire is in the other's magnetic field directed to produce attraction).
lF=2×10−4N/m, attractive.
Q.672Medium
For a uniform plane wave incident normally on a perfect conductor, the reflection coefficient for the electric field is:
Answer: C
At the surface of a perfect conductor (σ→∞, skin depth →0), the boundary condition requires the tangential electric field to be zero. The reflection coefficient for the electric field is
Γ=η2+η1η2−η1,
where η2=0 for a perfect conductor. Thus
Γ=0+η10−η1=−1.
The reflected electric field is equal in magnitude but opposite in phase to the incident field, resulting in a standing wave with a null at the conductor surface.
Q.673Medium
A parallel-plate capacitor has plates of area A=0.02m2 separated by a distance d=2mm filled with a dielectric of relative permittivity εr=5. The capacitance is:
Answer: B
The capacitance of a parallel-plate capacitor with a dielectric is