A DC shunt motor has an armature resistance of 0.5Ω and a field resistance of 200Ω. It is connected to a 220V supply. If the armature current at full load is 40A, the developed mechanical power is:
Answer: A
The back EMF is:
Eb=V−IaRa=220−(40×0.5)=220−20=200V
The developed mechanical power is:
Pmech=Eb×Ia=200×40=8000W
Q.342Medium
In a three-phase induction motor drive, the slip at maximum torque is denoted by sm. If the rotor resistance per phase is R2 and the rotor leakage reactance per phase at standstill is X2, then sm is given by:
Answer: A
From the equivalent circuit of an induction motor, differentiating the torque expression with respect to slip and setting it to zero gives the slip at maximum torque:
sm=X2R2
This shows that increasing rotor resistance (e.g., by external resistance in wound-rotor motors) increases sm, shifting the maximum torque to a higher slip value, which is the principle of rotor resistance speed control.
Q.343Medium
Which of the following methods of speed control of a DC shunt motor gives speed below the base (rated) speed?
Answer: A
The speed of a DC motor is given by N∝ϕVa−IaRa. Reducing the armature voltage Va below the rated value reduces the back EMF and hence the speed — this gives speeds below base speed at constant torque. Field weakening (reducing ϕ) gives speeds above base speed at constant power. Hence armature voltage control is used for sub-base speed operation.
Q.344Medium
In regenerative braking of a DC separately excited motor, the motor acts as a:
Answer: A
During regenerative braking, the kinetic energy of the load drives the motor above synchronous/no-load speed, causing the back EMF Eb to exceed the supply voltage Va. The machine then acts as a generator, and the current reverses direction, feeding electrical energy back into the DC supply. This is energy-efficient braking, unlike dynamic braking which dissipates energy as heat in resistors.
Q.345Medium
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.
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Q.346Medium
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.347Medium
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.348Medium
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.349Medium
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.
Q.350Medium
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.351Medium
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.352Medium
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.353Medium
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.354Medium
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.355Medium
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.356Medium
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.357Medium
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