ISRO HSFC Technician B Electrician Solved Question paper

Answer: "Same as "

In a transformer, the voltage is linked to the number of turns in the winding. The voltage per turn is the same for both the primary and secondary windings because they share the same magnetic core and the same magnetic flux.

For example:

• If 1 turn in the primary winding produces 2 volts, then 1 turn in the secondary winding will also produce 2 volts.
• The difference between the primary and secondary voltages comes from the number of turns in each winding.

In a step-up transformer, the secondary winding has more turns than the primary winding. This increases the total voltage in the secondary, but the voltage per turn remains constant for both windings.

So, voltage per turn in the primary = voltage per turn in the secondary.

Answer: "√2f "

An inductor’s impedance is twice that of a capacitor at a specific frequency (f).

Both components are connected in series.

We need to find the new resonant frequency.

Impedance relationship

The inductive reactance X_L and capacitive reactance X_C are related by:

X_L = 2X_C

Substituting their formulas:

$omega L = 2 cdot frac{1}{omega C}$

where

$ omega = 2pi f $ is the angular frequency.

Step 2: Solve for angular frequency $$ omega $$

Rearranging the equation:

$omega^2 L = frac{2}{C}$

$omega = sqrt{frac{2}{LC}}$

Step 3: Resonant frequency formula

The resonant angular frequency is related to the resonant frequency ($ f_text{new}$) by:

$omega = 2pi f_text{new}$

Substitute $ omega = sqrt{frac{2}{LC}}$:

$2pi f_text{new} = sqrt{frac{2}{LC}}$

Solve for $ f_text{new}$:

$f_text{new} = frac{sqrt{2}}{2pi sqrt{LC}}$

Step 4: Compare with the original resonant frequency

The original resonant frequency ($ f_text{original}$) is given by:

$f_text{original} = frac{1}{2pi sqrt{LC}}$

Now compare:

$f_text{new} = sqrt{2} cdot f_text{original}$

The new resonant frequency is:

$f_text{new} = sqrt{2}f$

Answer: "√3:1 "

Lambert’s Cosine Law

• Lambert’s cosine law states that the illumination (E) on a surface is directly proportional to the cosine of the angle (θ) between the direction of light and the normal to the surface. Mathematically, this can be expressed as:

E ∝ cos(θ)

Applying the Law to the Problem

Let’s denote:

• E₁: Illumination at 30 degrees
• E₂: Illumination at 60 degrees

Using Lambert’s cosine law:

E₁ ∝ cos(30°)

E₂ ∝ cos(60°)

To find the ratio of E₂ to E₁, we can write:

E₁/E₂ = cos(30°) / cos(60°)

Now, we know that:

• cos(60°) = 1/2
• cos(30°) = √3/2

Substituting these values:

E₁/E₂ = (√3/2)/(1/2)

Simplifying:

E₁/E₂ = √3/1

Therefore, the ratio of change in illumination is √3:1.

Answer: "0.99 mA "

Using the relationship between IC and IB:

IC = β * IB = 99 * IB

Where

β = DC current gain

IC = Collector current

IB = Base current

IC = 99IB

Applying KCL in loop 1:

-5 + 30(IB) + 0.7 + 4(IE) = 0

-5 + 30(IB) + 0.7 + 4(IB + IC) = 0

-5 + 30(IB) + 0.7 + 4(IB +99IB) = 0

-5 + 30(IB) + 0.7 + 4(100IB) = 0

430IB = 4.3

IB = 0.01 mA

IB = 10 μA

IC = 99IB

IC = 990 μA

or IC = 0.990 mA

Answer: "None of these "

Class A:

• Organic materials like cotton, silk, and paper when impregnated or immersed in oil.
• Enameled wire with enamel coating.
• Maximum operating temperature: 105°C (221°F).

Class B:

• Inorganic materials like mica, glass fiber, and asbestos (nowadays rarely used due to health concerns).
• May include organic binding substances in built-up form.
• Maximum operating temperature: 130°C (266°F).

Class E:

• Class E serves as an intermediate category between classes A and B.
• May include materials with slightly higher temperature ratings than Class A but not exceeding Class B.

Class F:

• Insulation based on mica, asbestos (rarely used), or glass fiber.
• Binding materials are high-temperature resistant, such as silicone.
• Maximum operating temperature: 155°C (311°F).

Class H:

• Similar to Class F but with even higher temperature resistance.
• Maximum operating temperature: 180°C (356°F).

वर्ग (Class)	Material	अधिकतम संचालन तापमान (Maximum Operating Temperature)
A	Cotton, silk, paper (impregnated or immersed in oil), enamel	105°C (221°F)
B	Mica, asbestos (अब कम इस्तेमाल किया जाता है), glass fiber	130°C (266°F)
E	Intermediate between A and B	105°C से 130°C के बीच (221°F से 266°F के बीच)
F	Mica, asbestos (rarely used), glass fiber with high-temperature resistant binding (e.g., silicone)	155°C (311°F)
H	Mica, asbestos (rarely used), glass fiber with even higher temperature resistant binding (e.g., silicone)	180°C (356°F)

Answer: "10 KHz "

The frequency of the signal at the clock of the rising edge triggered D flip flop is

f_out = 2 * f_in

where:

• f_out is the output frequency
• f_in is the input frequency

Therefore, if the output frequency (f_out) of the D flip-flop is 5 kHz, the input frequency (fin) of the clock signal must be twice the output frequency:

f_in = 2 * f_out = 2 * 5 kHz = 10 kH

Answer: "2V_m "

A positive clamper circuit is an electronic circuit designed to shift the entire waveform of an input signal upward, ensuring that the output voltage remains above a specified reference level, typically zero volts. This circuit is particularly useful in applications where it is necessary to prevent negative voltages from appearing at the output.

Positive peak detector circuit: This circuit detects the positive peak value of the input signal and provides it in the output.

This diagram can be broken out into two circuits:

1. Positive clamper circuit
2. Positive peak detector circuit

Vo = 2Vi

Vo = 2(Vm sinωt)

The peak value of output voltage is:

Vo = 2Vm

Answer: "10V "

Since the above circuit is an Inverting amplifier op-amp circuit, so the voltage gain will be:

Vo/Vi = -RF/R1

• Vo is the output voltage of the op-amp.
• Vi is the input voltage to the op-amp.
• RF is the feedback resistor connected between the output and the inverting input of the op-amp.
• R1 is the input resistor connected between the input voltage source and the inverting input of the op-amp.

Vo/3= -20/-5

Vo/3= 12

• The calculated ideal output voltage (12V) exceeds the positive saturation voltage (+10V) of the op-amp. This is an important point.
• Due to saturation, the op-amp cannot output a voltage beyond its saturation limits. In this case, even though the ideal output would be 12V, the actual output will be clamped to the positive saturation voltage (+10V).

Answer: "Cross Linked Paper ethylene cable "

XLPE stands for cross-linked polyethylene, which is a type of insulation used in electrical cables. It is particularly known for its excellent electrical, thermal, and mechanical properties, making it a popular choice for power transmission and distribution applications.

Key Features and Benefits of XLPE Cables

• High Electrical Strength: XLPE cables can withstand high voltage levels, making them suitable for both low and high voltage applications.
• Excellent Insulation Properties: They provide strong resistance to moisture, heat, and chemicals. This enhances their durability and performance in various environments.
• Mechanical Properties: XLPE cables exhibit good flexibility and are resistant to abrasion and impact, which is crucial for installations in challenging conditions.
• Lightweight Design: Compared to traditional cables, XLPE cables are lighter, making them easier to handle and install.
• Thermal Stability: XLPE can operate at higher temperatures (up to 90°C continuously and 140°C in emergencies) without degrading its insulating properties.
• Long Service Life: The lifespan of XLPE insulated cables can exceed 50 years under normal operating conditions, contributing to lower maintenance costs over time.
• Low Smoke Zero Halogen (LSZH): Many XLPE cables are designed to produce minimal smoke and no halogens when exposed to fire, enhancing safety in enclosed spaces.

Applications

XLPE cables are widely used in various sectors including:

• Power Distribution Networks: For both residential and industrial applications.
• High Voltage Transmission: Suitable for long-distance energy transmission.
• Outdoor Applications: Designed to withstand harsh environmental conditions.
• Mining and Construction: Due to their robust mechanical properties.

Answer: "Negative "

In an over compound d.c. generator, the voltage regulation is always Negative.

Voltage regulation = (no-load voltage – full load voltage)/full load voltage.

Over-Compounded Generator:

• The full-load terminal voltage is greater than the no-load voltage in an over-compounded generator. Series winding has positive coefficient (increasing current leads to increasing flux). which strengthens the main field with increasing load current, leading to higher induced voltage.
• This characteristic makes over-compounded generators suitable for applications requiring good voltage regulation even with varying loads, such as arc welding machines.

Flat-Compounded Generator:

• The flat-compounded generator, where the full-load and no-load terminal voltages are equal. This is achieved by carefully balancing the series and shunt field strengths, so the voltage increase from the series winding exactly compensates for the voltage drop due to armature resistance and reactance with increasing load.
• Flat-compounded generators are used in applications requiring constant voltage, such as lighting and battery charging.

Under-Compounded Generator:

• The under-compounded generator is correct. The terminal voltage decreases with increasing load due to the limited contribution of the series winding compared to the voltage drop in the armature.
• Under-compounded generators are not widely used due to their poor voltage regulation characteristics, but they can be found in specific applications where a slight voltage drop is acceptable or even desired.