Solution:

MCCB stands for Molded Case Circuit Breaker.

It is a type of electrical protection device which is used when the load current exceeds the limit of a miniature circuit breaker.

The MCCB provides protection against overload, short circuit faults and is also used for switching circuits.

Solution:

Electric power is the rate, per unit time, at which electrical energy is transferred by an electric circuit. The units of power is Watts. It can be expressed as

P = VI

Where V is the voltage

I is the current

Energy is the capacity to do work. Energy is power integrated over time. The units of energy are Joules.

$E = \mathop \smallint \nolimits_0^t {\rm{\;}}pdt$

Solution:

Armature windings are mainly of two types

1. Lap winding
2. Wave winding

A winding in which the number of parallel paths between the brushes is equal to the number of poles is called simplex lap winding.

In case of LAP wound DC Machines:

No. of parallel paths (A) = No. of poles (P) = 6

Solution:

Polystyrene is an example of an Insulator.

An insulator is a material that does not allows current to flow through it and having very high resistance.

Insulator has a negative temperature coefficient that means the temperature is increased resistance is decreased.

Examples of insulators are rubber, glass, wood, plastic, porcelain, Polystyrene, paper, etc.

Solution:

Power can define power as the rate of doing work, it is the work done in unit time.

Power is given by,

P = V × I

Calculation:

Given: I = 5e^−5t A

P = V × I

By exponential property,

e^t × e^t =e^2t 

P=50e^−10t W

Solution:

Direct Lighting:

• In the Direct lighting scheme, more than 90 percent of total light flux is made to fall directly on the working plane with the help of deep reflectors.
• This scheme is most efficient but causes hard shadows and glare.
• It is mainly used for industrial and general outdoor lighting.

Solution:

RMS values of an alternating voltage or current represent the real magnitude.

Root-Mean-Square (RMS) Value:

• The RMS value of an alternating current is given by that steady (DC) current which when flowing through a given circuit for a given time produces the same heat as produced by the alternating current when flowing through the same circuit for the same time.
• It is also known as the effective or virtual value of the alternating current, the former term being used more extensively.

RMS value of the alternating quantity is given as,

V_RMS = V_m/√2

I_RMS = I_m/√2

Solution:

A two-value capacitor-run motor starts with a High capacitor and runs with a Low capacitor.

Capacitor start capacitor run motors

• This type of motor is designed to operate with the start winding and it is a series capacitor permanently connected to the supply.
• lt has two capacitors connected in parallel in the start winding circuit for starting purposes.
• A start capacitor stays in the circuit long enough to rapidly bring the motor up to a predetermined speed, which is usually about 75% of the full speed, and is then taken out of the circuit, often by a centrifugal switch that releases at that speed.
• In a capacitor start and run the motor the function of the running capacitor in series with the auxiliary winding is to Improve power factor.
• Run capacitors are designed for continuous duty while the motor is powered, which is why electrolytic capacitors are avoided.
• As a result, this motor has a higher starting torque as well as a higher efficiency. These motors are employed in sizes ranging from 375 W to 7.5 kW.  The power factor of these motors is upto 80% to 100%.

Note:-

• Inductive Loads consume Reactive Power and Capacitive Loads deliver Reactive Power.
• The Power Factor Angle is inversely proportional to the Power Factor (reducing the angled gap between the Voltage vector and the Current Vector improved power factor).
• So in a single-phase Induction Motor, a Capacitor Bank is connected, then the Reactive power demand by the Inductive load (Induction Motor) is supplied by the Capacitor Bank, not from the Source.
• Therefore the system reactive power demand is reduced and the Power factor is improved.

Solution:

The angular velocity of a sinusoidal voltage is given as

ω = 2πf in radians / seconds

Where, f = frequency in Hz

Calculation:

Given, ω = 200π

200π  = 2 π × f

f = 100 Hz

Solution:

The capacitance between the conductors of a 1-ϕ, 2- long solid conductors transmission line is

= π ϵ_r ϵ_∘ / ln(D / r)

where

D = distance of separation between two long solid conductors

r = radius of each long solid conductor

Calculations:

ϵ∘ = 8.85 × 10^-12

ϵ_r  = 1 (for air)

π = 3.14

ln (D / r) = 2.303 log (D / r)

From equation ( i )

C_AB = (3.14 × 8.85 × 10-12 × 1) / (2.303 log (D / r))

C_AB = 0.0121/[ log (D/r)]  µF/km