This question tests the understanding of the condition for maximum efficiency in a transformer. For a transformer to operate at maximum efficiency, its iron losses (core losses) must be equal to its copper losses (ohmic losses).
D) 1 kW — At maximum efficiency, the iron loss (\(P_i\)) is equal to the copper loss (\(P_{cu}\)). Given \(P_i = 1 \text{ kW}\), therefore \(P_{cu} = 1 \text{ kW}\).
The question asks about the material used for electrodes in arc heating. Arc heating relies on creating an electric arc between two electrodes, which generates intense heat. The material chosen for these electrodes must possess specific properties to withstand the high temperatures and electrical discharge involved.
Correct Option: C) Graphite is widely used for electrodes in arc heating (e.g., electric arc furnaces) due to its extremely high sublimation point, good electrical conductivity, and relatively low cost.
The question asks about the common application of sodium lamps. To answer this, we need to recall the characteristics of sodium lamps and their typical uses in lighting.
B) Street Lights — Sodium lamps, particularly High-Pressure Sodium (HPS) lamps, are extensively used for street lighting, highway lighting, and other outdoor area lighting due to their high efficiency, long lifespan, and ability to provide good visibility in various weather conditions, despite their poor color rendering.
The question asks about the material used for a fuse wire. A fuse is a safety device that protects an electrical circuit from overcurrent. It works by melting and breaking the circuit when the current exceeds a safe level. Therefore, the material used for a fuse wire must have specific properties: low melting point and high resistivity.
Correct Option: B) Lead-tin alloy is used for fuse wires because it has a low melting point and suitable resistivity, allowing it to melt and break the circuit quickly when an overcurrent occurs, thus protecting electrical appliances and circuits.
The question asks about the number of primary modulation techniques. Modulation is the process of varying one or more properties of a high-frequency carrier wave with a low-frequency message signal. This process is essential for transmitting information over long distances.
Correct Option: C) 3. There are three primary types of modulation techniques: Amplitude Modulation (AM), Frequency Modulation (FM), and Phase Modulation (PM).
The question asks about the characteristics of frequency modulation (FM). We need to recall the definition and properties of FM to determine which statement is correct.
Let the carrier wave be represented by:
\[ c(t) = A_c \cos(2\pi f_c t) \]where \(A_c\) is the carrier amplitude and \(f_c\) is the carrier frequency.
Let the modulating signal be:
\[ m(t) = A_m \cos(2\pi f_m t) \]where \(A_m\) is the modulating signal amplitude and \(f_m\) is the modulating signal frequency.
In FM, the instantaneous frequency \(f_i\) of the modulated wave is given by:
\[ f_i = f_c + k_f m(t) \]where \(k_f\) is the frequency sensitivity. The instantaneous frequency varies with the modulating signal.
The FM wave is then given by:
\[ s(t) = A_c \cos\left(2\pi f_c t + 2\pi k_f \int m(t) dt\right) \]From this equation, it is clear that the amplitude of the FM wave remains \(A_c\), which is constant. The frequency term \(\left(2\pi f_c t + 2\pi k_f \int m(t) dt\right)\) varies according to the modulating signal.
A) amplitude of carrier wave remains the same. This is the defining characteristic of frequency modulation. The information is encoded in the variations of the carrier frequency, while its amplitude is kept constant to minimize noise interference.
The question asks about the role of mercury in a mercury arc rectifier. Understanding the basic operation of a mercury arc rectifier is key to identifying the correct function of mercury.
A) cathode — In a mercury arc rectifier, the pool of liquid mercury acts as the cathode. It is the negative electrode from which electrons are emitted to sustain the arc discharge, facilitating the rectification process.
To determine the AC to DC current ratio for a single-phase rotary converter operating at unity power factor, we need to consider the power balance between the AC input and DC output sides, assuming ideal conditions (no losses).
Understand Rotary Converters: A rotary converter is a type of electrical machine that converts AC power to DC power or vice versa. It combines the functions of a synchronous motor and a DC generator in a single armature winding.
Power Balance: For an ideal rotary converter, the input AC power equals the output DC power.
\[ P_{AC} = P_{DC} \]AC Power Input: For a single-phase AC input, the power is given by:
\[ P_{AC} = V_{AC} I_{AC} \cos\phi \]where \(V_{AC}\) is the RMS AC voltage, \(I_{AC}\) is the RMS AC current, and \(\cos\phi\) is the power factor.DC Power Output: The DC power output is given by:
\[ P_{DC} = V_{DC} I_{DC} \]where \(V_{DC}\) is the DC voltage and \(I_{DC}\) is the DC current.Unity Power Factor Condition: The problem states that the power factor is unity, so \(\cos\phi = 1\). Therefore, the AC power equation simplifies to:
\[ P_{AC} = V_{AC} I_{AC} \]Equating Powers: Setting \(P_{AC} = P_{DC}\):
\[ V_{AC} I_{AC} = V_{DC} I_{DC} \]Voltage Ratio for Single-Phase Rotary Converter: For a single-phase rotary converter, the theoretical relationship between the AC RMS voltage and the DC voltage is:
\[ V_{AC} = \frac{1}{\sqrt{2}} V_{DC} \]This relationship is derived from the fact that the AC voltage is the RMS value of the sinusoidal voltage induced in the armature, and the DC voltage is the average value of the rectified voltage.Substitute Voltage Ratio: Substitute the voltage ratio into the power balance equation:
\[ \left(\frac{1}{\sqrt{2}} V_{DC}\right) I_{AC} = V_{DC} I_{DC} \]Solve for Current Ratio: Cancel \(V_{DC}\) from both sides:
\[ \frac{1}{\sqrt{2}} I_{AC} = I_{DC} \]Rearrange to find the AC to DC current ratio \(\frac{I_{AC}}{I_{DC}}\):
\[ \frac{I_{AC}}{I_{DC}} = \sqrt{2} \]Calculate the Value:
\[ \sqrt{2} \approx 1.414 \]C) 1.4. The AC to DC current ratio for a single-phase rotary converter at unity power factor is \(\sqrt{2}\), which is approximately 1.414. This matches option C.
This question tests the fundamental definition of a superconductor. Superconductors are materials characterized by two key properties: zero electrical resistance and the expulsion of magnetic fields (Meissner effect) below a certain critical temperature.
Correct Option: B) zero
When a material becomes a superconductor, its electrical resistance, and consequently its resistivity, drops abruptly to zero below its critical temperature. This is a defining characteristic of superconductivity.
The problem asks us to identify the logic gate whose output is 1 when both inputs are 0. We need to evaluate the truth table for each type of gate mentioned in the options for the specific input condition (Input A = 0, Input B = 0).
Evaluate NAND gate: If inputs are A=0, B=0, then for an AND gate, output is \(0 \cdot 0 = 0\). A NAND gate is the inverse of an AND gate, so its output would be \(\overline{0 \cdot 0} = \overline{0} = 1\).
Evaluate XOR gate: If inputs are A=0, B=0, then for an XOR gate, output is \(0 \oplus 0 = 0\). The XOR operation is true if and only if inputs are different.
Evaluate OR gate: If inputs are A=0, B=0, then for an OR gate, output is \(0 + 0 = 0\).
Evaluate AND gate: If inputs are A=0, B=0, then for an AND gate, output is \(0 \cdot 0 = 0\).
Evaluate NOR gate: If inputs are A=0, B=0, then for an OR gate, output is \(0 + 0 = 0\). A NOR gate is the inverse of an OR gate, so its output would be \(\overline{0 + 0} = \overline{0} = 1\).
Evaluate Ex-NOR (XNOR) gate: If inputs are A=0, B=0, then for an XNOR gate, output is \(\overline{0 \oplus 0} = \overline{0} = 1\). The XNOR operation is true if and only if inputs are the same.
We are looking for a gate whose output is 1 when both inputs are 0. Based on our evaluation:
Now let's check the given options.
Option D states "NOR or Ex-NOR". As determined in steps 5 and 6, both NOR and Ex-NOR gates produce an output of 1 when both inputs are 0. Therefore, this option is correct.
A) NAND or XOR While a NAND gate gives an output of 1 for inputs (0,0), an XOR gate gives an output of 0 for inputs (0,0). So, this option is incorrect.
B) OR or XOR Both OR and XOR gates give an output of 0 for inputs (0,0). So, this option is incorrect.
C) AND or XOR Both AND and XOR gates give an output of 0 for inputs (0,0). So, this option is incorrect.