ISRO HSFC Technician B Electrician Solved Question paper

Answer: "1/7 "

A non-leap year has 365 days. Dividing 365 by 7 (days in a week), we get 52 weeks and 1 extra day.

Possible Outcomes for the Extra Day:

This extra day can be any of the 7 days of the week: Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, or Saturday.

Favorable Outcome:

For a non-leap year to have 53 Sundays, the extra day must be a Sunday.

Calculating Probability:

• Total Possible Outcomes: 7 (any day of the week)
• Favorable Outcome: 1 (Sunday)

Probability = Favorable Outcomes / Total Possible Outcomes

Therefore, the probability of having 53 Sundays in a non-leap year is:

Probability = 1/7

Answer: "Direction of emf "

Fleming’s right-hand rule is specifically used to determine the direction of induced current(emf) in generator’s windings conductor moving within a magnetic field.

Fleming’s right-hand rule:

1. Thumb: Represents the direction of the conductor’s motion relative to the magnetic field.
2. Index finger: Points in the direction of the magnetic field.
3. Middle finger: Points in the direction of the induced current.

So, by knowing the direction of the motion and the magnetic field, you can use this rule to determine the direction of the current flowing in the conductor.

Important note:

Although closely related, it’s important to differentiate between induced current and induced EMF.

• Induced EMF: Voltage generated in a conductor due to a changing magnetic field (Fleming’s right-hand rule applies to this).
• Induced current: The actual flow of electrons in the conductor due to the induced EMF. The direction of the induced current is always opposite to the direction of the induced EMF.

Therefore, while Fleming’s right-hand rule directly gives you the direction of the induced EMF, you can use this knowledge to understand the direction of the induced current by reversing it.

Finger	Right-hand rule	Left-hand rule
Thumb	The direction of motion of conductor (input)	The direction of the conductor (output)
Forefinger	Magnetic field	Magnetic field
Middle finger	The direction of induced emf (output)	The direction of current (input)

 

Answer: "210.4 V "

Given Information:

• Load current (IL) = 100 A
• Terminal voltage (Vt) = 200 V
• Shunt field resistance (Rsh) = 50 Ω
• Armature resistance (Ra) = 0.1 Ω

Step-by-Step Solution:

1. Calculate Shunt Field Current (Ish):

• • The shunt field current flows through the shunt field resistance.
  • Using Ohm’s law:
  • Ish = Vt / Rsh
  • Ish = 200 V / 50 Ω = 4 A

2. Calculate Armature Current (Ia):

• The armature current is the sum of the load current and the shunt field current.
• Ia = IL + Ish
• Ia = 100 A + 4 A = 104 A

3. Calculate Voltage Drop Across Armature Resistance (Va):

• Using Ohm’s law:
• Va = Ia * Ra
• Va = 104 A * 0.1 Ω = 10.4 V

4. Calculate Generated EMF (Eg):

• The generated EMF is the sum of the terminal voltage and the voltage drop across the armature resistance.
• Eg = Vt + Va
• Eg = 200 V + 10.4 V = 210.4 V

Therefore, the generated voltage (Eg) is 210.4 volts.

Answer: "Dropping "

Drooping Characteristics of DC Shunt Generators

DC shunt generators are known for their slightly drooping characteristics, which play a crucial role in their operation, especially when used in parallel configurations with other generators. This drooping behavior refers to the relationship between the terminal voltage and the load current, where the terminal voltage decreases slightly as the load current increases.

Key Characteristics

1. Voltage Regulation:

• Shunt generators maintain a relatively constant terminal voltage over a wide range of load currents due to their self-regulating nature. However, as the load increases, the terminal voltage exhibits a slight drop, characteristic of drooping behavior.

2. Parallel Operation:

• The slightly drooping characteristics make DC shunt generators particularly suitable for parallel operation. When multiple shunt generators are connected in parallel, they can share the load effectively. If one generator attempts to supply more than its share of the load, its terminal voltage will drop, which naturally reduces its output and helps maintain system stability.

3. Load Sharing:

• In parallel configurations, shunt generators will share the total load based on their voltage characteristics. If two generators have similar drooping characteristics, they will divide the load proportionally to their ratings. This ensures that each generator operates within its capacity without overloading any single unit[1][5].

4. Impact of Armature Reaction:

• The armature reaction affects the generated voltage by causing a drop due to increased armature current under load conditions. As the load increases, this effect becomes more pronounced, leading to a further reduction in terminal voltage.

Summary of Drooping Characteristics

• Slight Voltage Drop: Terminal voltage decreases as load current increases.
• Self-Regulation: Capable of maintaining nearly constant voltage across varying loads.
• Effective Parallel Operation: Facilitates stable and balanced load sharing among multiple generators.
• Armature Reaction Influence: Voltage drop is influenced by armature reaction effects at higher loads.

Overall, the drooping characteristics of DC shunt generators are essential for their functionality in power systems, particularly when ensuring stability and efficiency during parallel operations.

Answer: "1:π "

1. Circular Cross-Section:
   • The radius of the circular cross-section is denoted as “r”.
   • The area of this section, Ac, is calculated using the formula for the area of a circle:

Ac = πr²

• Square Cross-Section: The side length of the square cross-section is denoted as “a”.
• The area of this section, As, is calculated using the formula for the area of a square:

As = a²

2. Volume Conservation:

• • Since no material is lost during the reshaping process, and the length of both rods remains constant, their volumes must be equal.
  • Therefore, we can equate the volumes of the two rods:

Volume = Ac * L = As * L

 Calculate the Resistances:

• Resistance of the Circular Rod (Rc): The resistance of a conductor is given by the formula: R = ρL / A
• where:
• ρ is the resistivity of the material
• L is the length of the conductor
• A is the cross-sectional area

Therefore, the resistance of the circular rod is:

Rc = ρL / Ac = ρL / πr²

• Resistance of the Square Rod (Rs):

Similarly, the resistance of the square rod is:

Rs = ρL / As = ρL / r²

4. Ratio of Resistances:

• Now, we can find the ratio of the two resistances:
• Rc / Rs = (ρL / πr²) / (ρL / r²) = 1
• Therefore, the ratio of the resistances of the two rods is:

Rc : Rs = 1 : π

Answer: "Copper oxide diode "

A metal rectifier is an early type of semiconductor rectifier that uses a metal oxide as the semiconductor material.

Common types of metal rectifiers:

1. Copper Oxide Rectifiers: These use copper oxide as the semiconductor material.
2. Selenium Rectifiers: These use selenium as the semiconductor material.

How they work:

A metal rectifier consists of a metal plate (usually copper) coated with a thin layer of semiconductor material (copper oxide or selenium). When a voltage is applied across the device, current can flow in one direction (forward bias) but not in the other (reverse bias). This property is used to convert alternating current (AC) to direct current (DC).

• Gas tube rectifier: Uses a gas-filled tube and ionization to convert AC to DC.
• Silicon rectifier: Uses a semiconductor material (silicon) to convert AC to DC. This is the most common type of rectifier used today due to its small size, high efficiency, and low cost.

Answer: "Class C "

• In class C amplifiers, the transistor conducts for less than one-half cycle of the input signal (i.e. conduction angle is less than 180° around 80° to 120°).
• This reduced conduction angle improves the amplifier efficiency but creates a large amount of distortion.

Class	Class-A	Class-B	Class-C	Class-AB
Operating cycle	360°	180°	Less than 180°	More than 180° and less than 360°
Position of Q	Centre	On X-axis	Below X-axis	Above X- axis
Efficiency	25% or 50%	78.5%	Highest (almost 100 %)	50 % to 78.5 %
Distortion	Absent	Present more than class-A	Highest	Present less than class-B

Answer: "3A "

• The fusing current of any wire is approximate 1.5 to twice the current rating of the fuse.
• The value of fusing factor is always greater than 1.
• The small the value of fusing factor, greater is the chance of deterioration of fusing element due to overheating and oxidation.

Fusing current = Current rating * Fusing factor

40 SWG copper wire has a current rating of 1.5 A.

Fusing current = 2 × current rating

Fusing current = 3 A

Sr. No.	S.W.G.	Current Rating of Fuse (Amp)	Approx. Fusing Current (Amp)
1	40	1.5	3
2	39	2.5	4
3	38	3.0	5
4	37	3.5	6
5	36	4.5	7
6	35	5.5	9
7	34	6.0	10
8	33	7.0	11
9	32	8.0	12
10	31	9.0	13

Answer: "2.94 mA "

We know that,

Cut in voltage of Ge diode is .3V and of silicon is 0.7 V

The diode which has lower cut-in voltage will turned on first.

So, here Ge diode will conduct, and the current through 5 kΩ resistor is given by,

I = (15 – 0.3)/ 5 kΩ = 2.94 mA

Answer: "222 V "

When a magnetic flux linked with a coil changes, an electromotive force (EMF) is induced in the coil. This is Faraday’s Law of Electromagnetic Induction. The RMS value of the induced EMF in a coil is given by the formula:

E = 4.44 * Φₘ * f * N

Where:

• E = RMS value of the induced EMF
• Φₘ = Maximum flux linkage (in Webers)
• f = Frequency of the AC supply (in Hertz)
• N = Number of turns in the coil

Calculating the Induced EMF:

Substituting the given values into the formula:

E = 4.44 * 0.1 * 50 * 10

E = 222 V

Therefore, the RMS value of the induced EMF in the secondary is 222 volts.