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Past Questions
The depth of penetration of a wave in a lossy dielectric increases with increasing :
A)
Conductivity
B)
Permeability
C)
Wavelength
D)
Permittivity
The depth of penetration (also known as skin depth, δ) is the distance over which the amplitude of an electromagnetic wave decreases to 1/e (approximately 37%) of its original value as it penetrates a lossy medium.
For a good conductor or a lossy dielectric, the skin depth is given by the formula:
δ = 1 / α
In the case of a good conductor (where conductivity σ >> ωε), the formula simplifies to:
δ = √(2 / (ωμσ)) = √(1 / (π f μ σ))
Where:
- ω = angular frequency (2πf)
- μ = permeability
- σ = conductivity
- f = frequency
Analysis of the options:
1. Conductivity (σ): Since δ is inversely proportional to √σ, increasing conductivity decreases the depth of penetration.
2. Permeability (μ): Since δ is inversely proportional to √μ, increasing permeability decreases the depth of penetration.
3. Wavelength (λ): Wavelength is inversely proportional to frequency (λ = v/f). Since δ is inversely proportional to √f, it is directly proportional to √λ. Therefore, increasing the wavelength (which means decreasing the frequency) increases the depth of penetration.
4. Permittivity (ε): In a lossy dielectric, the attenuation constant α depends on permittivity. Generally, increasing permittivity increases the attenuation, thereby decreasing the penetration depth.
Therefore, the depth of penetration increases as the wavelength increases.
What is the duty cycle of the output of an astable multivibrator?
A)
50%
B)
100%
C)
75%
D)
55%
An astable multivibrator is a free-running oscillator that switches between two states without any external trigger. The duty cycle is defined as the ratio of the 'ON' time (T_on) to the total period (T = T_on + T_off). In a standard, symmetric astable multivibrator circuit (where the RC time constants for both charging and discharging paths are identical), the time spent in the high state is equal to the time spent in the low state. Therefore, T_on = T_off, and the duty cycle is (T_on / (T_on + T_off)) * 100% = (T_on / 2T_on) * 100% = 50%. While it is possible to design an astable multivibrator with a variable duty cycle by using different resistor or capacitor values, in the context of general theoretical exam questions where no specific component values are provided, the 'ideal' or 'standard' astable multivibrator is assumed to be symmetric, resulting in a 50% duty cycle.
How many select lines are contained in a multiplexer with 1024 inputs and one output?
A)
512
B)
258
C)
64
D)
10
A Multiplexer (MUX) is a combinational circuit that selects one of many input signals and forwards the selected input into a single line. The relationship between the number of input lines (n) and the number of select lines (m) is defined by the formula: n = 2^m, where 'n' is the number of inputs and 'm' is the number of select lines.
Given:
Number of inputs (n) = 1024
We need to find 'm' such that:
2^m = 1024
Calculating the powers of 2:
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256
2^9 = 512
2^10 = 1024
Therefore, m = 10.
Analysis of options:
- Option (a) 512: Incorrect. This is 2^9, not the exponent.
- Option (b) 258: Incorrect. This has no mathematical relationship to the power of 2 for 1024.
- Option (c) 64: Incorrect. This is 2^6.
- Option (d) 10: Correct. 2 raised to the power of 10 equals 1024.
If input frequency is 50Hz then ripple frequency of center tapped full wave rectifier will be equal to :
A)
100Hz
B)
50Hz
C)
25Hz
D)
500Hz
1. Understanding the Rectifier Type: A center-tapped full-wave rectifier utilizes two diodes to convert both the positive and negative half-cycles of the AC input into a unidirectional (DC) output.
2. Analysis of the Waveform: In a half-wave rectifier, only one half-cycle is passed, so the output frequency equals the input frequency. However, in a full-wave rectifier, the negative half-cycle is 'flipped' to become positive. This means for every one cycle of the input AC signal, there are two pulses (peaks) in the output DC signal.
3. Mathematical Calculation: The ripple frequency (f_ripple) for a full-wave rectifier is given by the formula: f_ripple = 2 × f_input. Given that the input frequency (f_input) is 50Hz, the ripple frequency is 2 × 50Hz = 100Hz.
4. Evaluation of Options:
- Option (a) 100Hz is correct based on the full-wave rectification principle.
- Option (b) 50Hz would be correct for a half-wave rectifier.
- Option (c) 25Hz and (d) 500Hz are mathematically incorrect for this circuit configuration.
What is the use of the compensation capacitor in op-amp?
A)
Improves the amplification of op-amp
B)
Decreases the slew rate of op-amp
C)
Increases the bandwidth of op-amp
D)
Op-amp acts as all pass filter
In an operational amplifier (op-amp), a compensation capacitor is primarily used to ensure stability and prevent oscillations by introducing a dominant pole in the frequency response. This process is known as frequency compensation.
1. **Stability vs. Slew Rate**: The compensation capacitor (typically placed in the internal gain stage) limits the rate at which the output voltage can change. The Slew Rate (SR) is defined as the maximum rate of change of output voltage per unit time. Since the capacitor must be charged and discharged by the internal current sources, adding a larger compensation capacitor increases the time required for the output to change, thereby decreasing the slew rate.
2. **Why other options are incorrect**:
- **Option a**: Compensation capacitors do not improve amplification; in fact, they are used to intentionally limit the gain at higher frequencies to maintain stability.
- **Option c**: Compensation capacitors actually *decrease* the bandwidth. By creating a dominant pole at a very low frequency, the unity-gain bandwidth is reduced to ensure the phase shift does not reach 180 degrees while the gain is still $\ge 1$ (maintaining a positive phase margin).
- **Option d**: An op-amp with a compensation capacitor does not act as an all-pass filter; it acts as a low-pass filter in terms of its open-loop gain response.
Which one of the following is a wideband amplifier?
A)
RF amplifier
B)
IF amplifier
C)
Video amplifier
D)
AF amplifier
A wideband amplifier is one that can amplify a broad range of frequencies without significant attenuation or phase distortion.
1. **Video Amplifier (Option c):** Video signals contain a very wide range of frequencies (from near DC up to several MHz) to maintain the detail and sharpness of the image. Therefore, video amplifiers are specifically designed as wideband amplifiers to handle this extensive bandwidth.
2. **AF (Audio Frequency) Amplifier (Option d):** These operate in the range of 20 Hz to 20 kHz. While essential, this range is very narrow compared to video or RF signals.
3. **IF (Intermediate Frequency) Amplifier (Option b):** These are designed to amplify a very specific, narrow band of frequencies centered around the intermediate frequency (e.g., 455 kHz for AM radio) to ensure selectivity. They are narrow-band amplifiers.
4. **RF (Radio Frequency) Amplifier (Option a):** While they operate at high frequencies, they are typically designed to amplify a specific carrier frequency or a narrow band around it to avoid interference from adjacent channels. They are generally considered narrow-band compared to video amplifiers.
To increase the bandwidth of a control system, we may use :
A)
Phase lag compensator
B)
Phase lead compensator
C)
Phase leg-lead compensator
D)
All of these
Bandwidth in a control system is directly related to the speed of response; a higher bandwidth implies a faster response and a shorter rise time.
1. **Phase Lead Compensator**: A lead compensator adds a positive phase angle to the system. This shifts the root locus to the left in the s-plane, which increases the system's natural frequency and damping. Consequently, it increases the gain-crossover frequency, which directly increases the bandwidth of the system. This results in a faster transient response.
2. **Phase Lag Compensator**: A lag compensator is primarily used to improve the steady-state accuracy (reduce steady-state error) by increasing the low-frequency gain. However, it typically reduces the bandwidth and slows down the transient response of the system.
3. **Lag-Lead Compensator**: While a lag-lead compensator combines both effects, its primary purpose is to achieve both steady-state accuracy (via the lag part) and improved transient response (via the lead part). However, if the specific goal is simply to 'increase' the bandwidth, the lead compensator is the primary tool used for this purpose.
Therefore, to increase the bandwidth and speed up the system, a Phase Lead Compensator is used.
To construct a 512 K x 8 memory , the number of 32 K x 4 memory circuit required is :
A)
16
B)
32
C)
8
D)
64
To determine the number of smaller memory chips required to construct a larger memory array, we compare the total capacity of the target memory with the capacity of a single available chip.
1. **Target Memory Capacity**:
The target memory is 512 K x 8.
Total bits = (512 * 1024) words * 8 bits/word = 4,194,304 bits.
2. **Available Chip Capacity**:
The available chip is 32 K x 4.
Total bits = (32 * 1024) words * 4 bits/word = 131,072 bits.
3. **Calculation of Number of Chips**:
Number of chips = (Total Target Capacity) / (Capacity of one chip)
Number of chips = (512 K * 8) / (32 K * 4)
Number of chips = (512 / 32) * (8 / 4)
Number of chips = 16 * 2 = 32.
Alternatively, looking at it by dimensions:
- To increase the word capacity from 32K to 512K, we need 512/32 = 16 rows of chips.
- To increase the word size (bit width) from 4 bits to 8 bits, we need 8/4 = 2 columns of chips.
- Total chips = 16 rows * 2 columns = 32 chips.
Which one of the following is not correct Maxwell equation?
A)
∇ × H = J + dD/dt
B)
∇ × E = -dB/dt
C)
∇ × D = ρ
D)
∇ × B = 0
To identify the incorrect Maxwell equation, let's review the four fundamental Maxwell's equations in differential form:
1. Gauss's Law for Electricity: ∇ · D = ρ (The divergence of the electric flux density is equal to the volume charge density).
2. Gauss's Law for Magnetism: ∇ · B = 0 (The divergence of the magnetic flux density is zero, meaning there are no magnetic monopoles).
3. Faraday's Law of Induction: ∇ × E = -∂B/∂t (The curl of the electric field is equal to the negative rate of change of the magnetic flux density).
4. Ampere-Maxwell Law: ∇ × H = J + ∂D/∂t (The curl of the magnetic field intensity is equal to the sum of the conduction current density and the displacement current density).
Evaluating the options:
- Option (a): ∇ × H = J + dD/dt is the correct Ampere-Maxwell Law.
- Option (b): ∇ × E = -dB/dt is the correct Faraday's Law.
- Option (c): ∇ × D = ρ is INCORRECT. The correct equation is ∇ · D = ρ (Divergence, not Curl). The curl of D is not fundamentally defined as equal to charge density.
- Option (d): ∇ × B = 0 is not one of the four primary Maxwell equations in general form (the primary one is ∇ · B = 0). However, in the context of this specific multiple-choice question, Option (c) is a blatant mathematical error regarding the operator (Curl vs Divergence) for a fundamental law, making it the most 'incorrect' choice. Note: In static magnetic fields, ∇ × B = μ₀J, but ∇ · B = 0 is the universal law. Comparing (c) and (d), (c) replaces a divergence operator with a curl operator for Gauss's Law, which is a fundamental conceptual error.
Find the reflection coefficient of the wave with SWR of 3.5.
A)
0.55
B)
0.23
C)
0.48
D)
0.68
Reflection coefficient |Γ| = (VSWR-1)/(VSWR+1) = (3.5-1)/(3.5+1)=2.5/4.5≈0.55.
Which one of the following is not a closed loop system?
A)
Respiratory system of an animal
B)
Execution of a program by computer
C)
Air condition system
D)
Driving a car
A closed-loop system (or feedback control system) is one where the output has a direct effect on the control action, typically through a feedback loop that compares the actual output with a desired reference to reduce error.
1. Respiratory system of an animal (Option a): This is a closed-loop system. The body monitors CO2 levels in the blood (feedback); if levels rise, the brain signals the lungs to increase the breathing rate to return to homeostasis.
2. Air condition system (Option c): This is a classic closed-loop system. A thermostat measures the room temperature (feedback) and compares it to the set point. The AC turns on or off based on this error signal to maintain the desired temperature.
3. Driving a car (Option d): This is a closed-loop system. The driver sees the position of the car relative to the lane (feedback) and continuously adjusts the steering wheel to keep the car on track.
4. Execution of a program by computer (Option b): In its basic form, the execution of a sequence of instructions in a program is an open-loop process. The CPU fetches an instruction and executes it regardless of the 'output' or result of that specific operation on the external environment. While high-level software may implement feedback loops, the fundamental act of executing a program (instruction cycle) is a linear, feed-forward process.
A full-adder can be made out of :
A)
Two half -adders
B)
Two half-adders and a NOT gate
C)
Two half-adders and a AND gate
D)
Two half-adders and a OR gate
A Full-Adder is a combinational circuit that adds three binary bits: two significant bits (A and B) and a carry bit (Cin) from a previous stage.
1. **Half-Adder (HA) Logic**: A half-adder takes two inputs and produces two outputs: Sum = A ⊕ B and Carry = A · B.
2. **Construction of Full-Adder using Half-Adders**:
- **First Half-Adder**: Takes inputs A and B. It produces a partial sum (S1 = A ⊕ B) and a partial carry (C1 = A · B).
- **Second Half-Adder**: Takes the partial sum (S1) and the carry-in (Cin) as inputs. It produces the final Sum (S = S1 ⊕ Cin) and another partial carry (C2 = S1 · Cin).
- **Final Carry Logic**: The final carry-out (Cout) of the full-adder is true if either the first half-adder produced a carry OR the second half-adder produced a carry. Therefore, Cout = C1 OR C2.
3. **Conclusion**: To implement a full-adder, we need two half-adders to handle the summation logic and one OR gate to combine the carry signals.
- Option (a) is incorrect because two half-adders alone cannot combine the carry signals.
- Option (b) and (c) are incorrect because NOT and AND gates do not perform the required carry-combination logic.
The circuit that will work as OR gate in positive level, will work as _______ gate in negative level logic.
A)
NOR gate
B)
NAND gate
C)
Both NAND and NOR gate
D)
AND gate
To solve this, we must understand the difference between Positive Logic and Negative Logic.
1. **Positive Logic**: Logic '1' is represented by a High voltage (Vcc) and Logic '0' is represented by a Low voltage (Ground).
2. **Negative Logic**: Logic '1' is represented by a Low voltage (Ground) and Logic '0' is represented by a High voltage (Vcc).
Let's analyze the OR gate in Positive Logic:
- Inputs (A, B) | Output (Y)
- 0, 0 -> 0
- 0, 1 -> 1
- 1, 0 -> 1
- 1, 1 -> 1
Now, we translate these voltage levels into Negative Logic. In Negative Logic, the binary values are inverted (0 becomes 1, and 1 becomes 0):
- Positive Logic (0, 0) is Negative Logic (1, 1). Output 0 becomes 1.
- Positive Logic (0, 1) is Negative Logic (1, 0). Output 1 becomes 0.
- Positive Logic (1, 0) is Negative Logic (0, 1). Output 1 becomes 0.
- Positive Logic (1, 1) is Negative Logic (0, 0). Output 1 becomes 0.
Looking at the Negative Logic truth table:
- Inputs (A, B) | Output (Y)
- 1, 1 -> 1
- 1, 0 -> 0
- 0, 1 -> 0
- 0, 0 -> 0
This truth table (where the output is '1' only when both inputs are '1') is the definition of an AND gate.
Alternatively, using De Morgan's Law: An OR gate in positive logic is equivalent to an AND gate in negative logic, and an AND gate in positive logic is equivalent to an OR gate in negative logic.
The unit of ∇ × H is :
A)
Ampere
B)
Ampere/meter
C)
Ampere/meter2
D)
Ampere-meter
Step-by-step reasoning:
1. Identify the variables: H represents the Magnetic Field Intensity. The SI unit of H is Ampere per meter (A/m).
2. Identify the operator: The symbol ∇ × represents the 'curl' operator. In Cartesian coordinates, the curl involves taking partial derivatives with respect to spatial coordinates (x, y, z).
3. Analyze the units of the curl operation: The curl of a vector field is defined as the derivative of the field components divided by a length element (e.g., ∂Hy/∂x).
4. Calculate the resulting unit:
Unit of (∇ × H) = (Unit of H) / (Unit of length)
Unit of (∇ × H) = (A/m) / m = A/m².
5. Physical Context: According to Ampere's Law in differential form, ∇ × H = J, where J is the current density. The unit of current density (J) is current per unit area, which is Ampere per square meter (A/m²). Therefore, the unit of ∇ × H must be A/m².
Comparison with options:
- Option (a) Ampere: This is the unit of current (I), not current density.
- Option (b) Ampere/meter: This is the unit of H itself.
- Option (c) Ampere/meter²: This is the correct unit for current density (J) and the curl of H.
- Option (d) Ampere-meter: This is the unit of magnetic pole strength or the integral of H over a path.
A plane EM wave at an oblique incidence at Brewster angle then the reflected wave will be :
A)
Purely parallel polarized
B)
Purely perpendicular polarized
C)
Both
D)
Neither of the two
1. Brewster's Angle Definition: Brewster's angle ($ heta_B$) is the specific angle of incidence at which light (or an EM wave) with a particular polarization is perfectly transmitted through a transparent dielectric surface with zero reflection.
2. Parallel Polarization (p-polarization): When an EM wave is polarized parallel to the plane of incidence, and it strikes the interface at the Brewster angle, the reflection coefficient for this polarization becomes zero. This means the reflected wave contains no parallel polarized component.
3. Perpendicular Polarization (s-polarization): The Brewster effect only applies to the parallel component. The perpendicular component (polarized perpendicular to the plane of incidence) is always partially reflected, regardless of whether the angle is the Brewster angle or not.
4. Conclusion: If a general plane EM wave (containing both parallel and perpendicular components) hits a surface at the Brewster angle, the parallel component is completely transmitted, and only the perpendicular component is reflected. Therefore, the reflected wave is purely perpendicular polarized.
Voltage series feedback (also called series shunt feed-back) results in :
A)
Increase in both input and output impedance
B)
Decrease in both input and output impedance
C)
Increase in input impedance and decrease in output impedance
D)
Decrease in input impedance and increase in output impedance
In feedback amplifiers, the effect on input and output impedance depends on how the feedback signal is connected (series or shunt) at the input and output ports.
1. **Input Impedance (Series Connection):** When the feedback is connected in 'series' with the input signal, the feedback voltage opposes the input voltage. This effectively increases the input impedance of the amplifier. Mathematically, Z_in(f) = Z_in(no feedback) * (1 + Aβ), where A is the open-loop gain and β is the feedback factor.
2. **Output Impedance (Shunt Connection):** When the feedback is connected in 'shunt' (parallel) at the output, the feedback network draws a portion of the output current. This results in a decrease in the output impedance. Mathematically, Z_out(f) = Z_out(no feedback) / (1 + Aβ).
Therefore, a 'Voltage Series Feedback' (also known as Series-Shunt feedback) increases the input impedance and decreases the output impedance. This configuration is ideal for creating a voltage amplifier, as it provides a high input impedance (to avoid loading the source) and a low output impedance (to ensure maximum voltage transfer to the load).
A bulb in a staircase has two switches, one switch being at the ground floor and the other one at the first floor. The bulb can be turned ON and also can be turned OFF by and one of the switches irrespective of the state of the other switch. The logic of switching of the bulb resembles.
A)
an AND gate
B)
an OR gate
C)
an XOR gate
D)
a NAND gate
Step-by-step reasoning:
1. **Analyze the Requirement**: The problem describes a standard staircase wiring system. The key requirement is that the state of the bulb (ON/OFF) can be changed by toggling either switch, regardless of the position of the other switch.
2. **Truth Table Construction**:
Let S1 be the switch at the ground floor and S2 be the switch at the first floor. Let L be the state of the bulb (0 = OFF, 1 = ON).
- If both switches are in the same position (both DOWN or both UP), let's assume the light is OFF: (0,0) -> 0 and (1,1) -> 0.
- If the switches are in opposite positions, the circuit is completed and the light is ON: (0,1) -> 1 and (1,0) -> 1.
3. **Matching with Logic Gates**:
- **AND gate**: Output is 1 only if both inputs are 1. (Does not match).
- **OR gate**: Output is 1 if any input is 1. (Does not match, as (1,1) would be 1).
- **XOR (Exclusive OR) gate**: Output is 1 if and only if the inputs are different. This perfectly matches the truth table: 0 XOR 0 = 0; 0 XOR 1 = 1; 1 XOR 0 = 1; 1 XOR 1 = 0.
- **NAND gate**: Output is 0 only if both inputs are 1. (Does not match).
4. **Conclusion**: The behavior where changing the state of any single input flips the output state is the defining characteristic of an XOR gate (or an XNOR gate, but XOR is the standard representation for this logic in multiple-choice questions of this type).
What is the characteristic of a good control system?
A)
Insensitive to the parameter variation but sensitive to the input commands
B)
Neither sensitive to parameter variations nor sensitive to input commands
C)
Insensitive to the input command
D)
Sensitive to parameter variation
In control system engineering, a 'good' or 'ideal' control system is characterized by its ability to track a desired input accurately while remaining robust against internal changes or external disturbances.
1. **Sensitivity to Input Commands**: A control system must be highly sensitive to the input (reference signal). This ensures that the system responds quickly and accurately to the user's commands, minimizing the steady-state error between the desired output and the actual output.
2. **Insensitivity to Parameter Variation**: Parameters of a system (such as resistance, capacitance, or mass) can change over time due to aging, temperature fluctuations, or manufacturing tolerances. A good system should be 'robust,' meaning its performance (stability and accuracy) should not degrade significantly when these internal parameters vary.
Evaluating the options:
- Option (a) correctly identifies both requirements: high sensitivity to the command and low sensitivity (insensitivity) to parameter changes.
- Option (b) is incorrect because a system that is not sensitive to input commands would not follow the desired setpoint.
- Option (c) is incorrect for the same reason as (b).
- Option (d) is incorrect because sensitivity to parameter variation leads to instability or inconsistent performance.
The memory technology which needs the least power is :
A)
ECL
B)
TTL
C)
MOS
D)
CMOS
To determine which technology consumes the least power, we must analyze the power dissipation characteristics of each logic family:
1. **ECL (Emitter-Coupled Logic):** This is the fastest logic family but consumes the most power because its transistors operate in the active region and draw constant current regardless of the state.
2. **TTL (Transistor-Transistor Logic):** Uses Bipolar Junction Transistors (BJTs). It consumes significantly less power than ECL but much more than MOS-based technologies because it requires constant current to maintain logic levels.
3. **MOS (Metal-Oxide-Semiconductor):** Specifically referring to NMOS or PMOS. While they consume less power than TTL, they often have static power dissipation when the output is low (in NMOS).
4. **CMOS (Complementary Metal-Oxide-Semiconductor):** This technology uses both p-type and n-type MOSFETs in complementary pairs. The key advantage is that in a steady state (logic 0 or 1), one of the transistors is always OFF, meaning there is virtually no static current flow from the supply to the ground. Power is consumed primarily during switching (dynamic power). Therefore, CMOS has the lowest overall power consumption.
Conclusion: CMOS is the industry standard for low-power memory (like SRAM and DRAM) and microprocessors due to its minimal static power dissipation.
If a system is given unbounded input then the system is :
A)
Stable
B)
Unstable
C)
Not defined
D)
Linear
In control systems, stability is fundamentally defined by the system's response to an input. According to the Bounded-Input Bounded-Output (BIBO) stability criterion, a system is considered stable if every bounded input results in a bounded output. Conversely, if a system produces an unbounded output for a bounded input, it is unstable. However, the question asks about the state of the system when given an 'unbounded input'. By definition, if the input itself is unbounded (tends toward infinity), the output of a causal linear time-invariant (LTI) system will also be unbounded. In the context of standard engineering examination terminology for this specific question, a system that cannot maintain a bounded output (which is inevitable when the input is unbounded) is categorized as unstable.
Breakdown of options:
- a. Stable: A stable system requires bounded inputs to produce bounded outputs; it cannot 'stabilize' an unbounded input into a bounded output.
- b. Unstable: Correct, as the system response will grow without bound.
- c. Not defined: The behavior is well-defined by the system's transfer function.
- d. Linear: Linearity is a property of the system's structure, not a result of the type of input provided.
What is the role of emitter resistance in the transistor amplifying circuit?
A)
To prevent thermal runaway
B)
To prevent increase in gain
C)
To lower the output impedance
D)
To increase gain
In a transistor amplifying circuit, an emitter resistor (Re) is used to provide negative feedback, which stabilizes the operating point (Q-point) against variations in temperature and transistor parameters (like beta).
1. **Mechanism**: As temperature increases, the collector current (Ic) tends to increase. This increase in Ic leads to an increase in the voltage drop across the emitter resistor (Ve = Ic * Re).
2. **Feedback**: Since the base-emitter voltage (Vbe) is equal to Vb - Ve, an increase in Ve causes Vbe to decrease.
3. **Stabilization**: A decrease in Vbe reduces the base current (Ib), which in turn reduces the collector current (Ic), thereby counteracting the initial increase caused by temperature.
4. **Thermal Runaway**: Without this stabilization, an increase in temperature leads to more current, which generates more heat, further increasing the temperature in a positive feedback loop until the transistor is destroyed. This process is called thermal runaway. Therefore, the emitter resistor prevents it.
Regarding other options:
- Option b: While emitter resistance does decrease the voltage gain (due to degeneration), its primary 'role' or purpose in circuit design is stability, not simply to 'prevent increase in gain'.
- Option c: Emitter resistance typically increases the input impedance and does not inherently lower the output impedance.
- Option d: Emitter resistance actually decreases the voltage gain (Gain ≈ Rc / (re + Re)), so it does not increase gain.
A basic S-R flip-flop can be constructed by cross-coupling of which basic logic gates?
A)
AND or OR gates
B)
XOR or XNOR gates
C)
NOR or NAND gates
D)
AND or NOR gates
A Set-Reset (S-R) flip-flop is the most basic sequential logic circuit. It is constructed using two cross-coupled logic gates, where the output of one gate is fed back as an input to the other.
1. **NOR Gate Implementation**: In a NOR-based SR flip-flop, the inputs are active-high. When the 'Set' input is high, the output Q becomes high. When the 'Reset' input is high, the output Q becomes low.
2. **NAND Gate Implementation**: In a NAND-based SR flip-flop (often called an $\overline{S}$-$\overline{R}$ latch), the inputs are active-low. The cross-coupling of NAND gates creates a stable state that holds the bit value until a change is triggered by the inputs.
Other options are incorrect because:
- **AND/OR gates** (Option a and d) are combinational gates that do not possess the inverting property required to create the regenerative feedback loop necessary for memory (latching).
- **XOR/XNOR gates** (Option b) are used for parity checking or arithmetic operations (like half-adders) and cannot be cross-coupled to form a stable SR latch.
The incoming solar radiation at a place on the surface of the earth is 1.2 KW/m2 . The amplitude of the electric field corresponding to this incident power is nearly equal to :
A)
80 mV / m
B)
2.5 V / m
C)
30 V /m
D)
950 V / m
To find the amplitude of the electric field from the incident solar radiation (power density), we use the formula for the intensity (Poynting vector magnitude) of an electromagnetic wave in free space:
1. **Formula**: The average power density (Intensity, I) is given by:
I = (1/2) * ε₀ * c * E₀²
Where:
- I = 1.2 kW/m² = 1200 W/m²
- ε₀ (Permittivity of free space) ≈ 8.854 × 10⁻¹² F/m
- c (Speed of light) ≈ 3 × 10⁸ m/s
- E₀ = Amplitude of the electric field (V/m)
2. **Alternative Formula**: Using the intrinsic impedance of free space (η₀ ≈ 377 Ω or 120π Ω):
I = E₀² / (2 * η₀)
3. **Calculation**:
E₀² = 2 * η₀ * I
E₀² = 2 * 377 * 1200
E₀² = 904,800
E₀ = √904,800
E₀ ≈ 951.2 V/m
4. **Conclusion**: The calculated value of approximately 951 V/m aligns most closely with option (d) 950 V/m.
- Option (a) 80 mV/m is far too low.
- Option (b) 2.5 V/m is far too low.
- Option (c) 30 V/m is far too low.
The operational amplifier is basically a :
A)
Low gain ac amplifier
B)
High gain dc amplifier
C)
High gain RC coupled amplifier
D)
Low gain transformer-coupled amplifier
An Operational Amplifier (Op-Amp) is defined as a high-gain electronic voltage amplifier with a differential input and a single-ended output.
1. **High Gain**: In an open-loop configuration, an ideal op-amp has infinite gain, and practical op-amps have very high open-loop gains (typically $10^5$ to $10^6$). This allows them to be used in feedback configurations to achieve precise closed-loop gains.
2. **DC Amplifier**: Op-amps are designed to amplify signals starting from 0 Hz (DC). They are direct-coupled amplifiers, meaning they do not use coupling capacitors (like RC coupling) that would block DC components. This makes them suitable for integrating, differentiating, and summing DC and AC signals.
Why other options are incorrect:
- **Option a**: Op-amps are characterized by high gain, not low gain.
- **Option c**: Op-amps are direct-coupled, not RC coupled. RC coupling is used in audio amplifiers to block DC bias, whereas op-amps need to process DC.
- **Option d**: Op-amps are high gain and typically use direct coupling rather than transformer coupling.
Two thin parallel wires are carrying current in the same direction. The force experienced between them is :
A)
Repulsive
B)
No force exists
C)
Attractive
D)
Perpendicular to their axis joining wires
The force between two parallel current-carrying conductors can be explained using Ampere's Law and the Lorentz Force Law.
1. **Magnetic Field Generation**: According to Ampere's Law, a current-carrying wire produces a magnetic field around it. For a wire carrying current I₁, the magnetic field (B₁) at the location of the second wire is circular and perpendicular to the wire.
2. **Force Calculation**: The force (F) experienced by the second wire carrying current I₂ in the presence of the magnetic field B₁ is given by the formula F = I₂ (L × B₁), where L is the length of the wire.
3. **Direction (Right-Hand Rule)**:
- If the currents are in the **same direction**, the magnetic field produced by the first wire interacts with the current of the second wire such that the resulting force is directed toward the first wire.
- Conversely, if the currents are in opposite directions, the force is directed away from the other wire.
Therefore, parallel currents flowing in the same direction attract each other, while currents flowing in opposite directions repel each other.
- Option (a) is incorrect because repulsion occurs with opposite currents.
- Option (b) is incorrect because a magnetic field is present, thus a force must exist.
- Option (d) is incorrect because the force acts along the line joining the two wires, not perpendicular to it.