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For a transmission line, the distributed parameters is/are ____________
For a transmission line, the distributed parameters are inductance per unit length (L), capacitance per unit length (C), resistance per unit length (R), and conductance per unit length (G).
For a transmission line, the distributed parameters are inductance per unit length (L), capacitance per unit length (C), resistance per unit length (R), and conductance per unit length (G).
See lessBundle conductor ______ inductance and ______ capacitance of a transmission line.
Bundle conductor reduces inductance and increases capacitance of a transmission line.
Bundle conductor reduces inductance and increases capacitance of a transmission line.
See lessBundled conductors used in the EHV transmission lines result into __________
Bundled conductors used in the EHV transmission lines result in reduced corona loss and improved electrical performance.
Bundled conductors used in the EHV transmission lines result in reduced corona loss and improved electrical performance.
See lessIn the p & n regions of the p-n junction the _________ & the ___________ are the minority charge carriers respectively.
In the p-n junction, the electrons are the minority charge carriers in the p-region, and the holes are the minority charge carriers in the n-region.
In the p-n junction, the electrons are the minority charge carriers in the p-region, and the holes are the minority charge carriers in the n-region.
See lessIn the p & n regions of the p-n junction the _________ & the ___________ are the majority charge carriers respectively.
In the p-n junction, the holes and the electrons are the majority charge carriers respectively.
In the p-n junction, the holes and the electrons are the majority charge carriers respectively.
See lessThe n-region has a greater concentration of _________ as compared to the p- region in a P-N junction diode.
The n-region has a greater concentration of electrons as compared to the p-region in a P-N junction diode.
The n-region has a greater concentration of electrons as compared to the p-region in a P-N junction diode.
See lessIn the p & n regions of the p-n junction the _________ & the ___________ are the majority charge carriers respectively.
In the p & n regions of the p-n junction, the holes and the electrons are the majority charge carriers respectively.
In the p & n regions of the p-n junction, the holes and the electrons are the majority charge carriers respectively.
See lessA p-type semiconductor material is doped with ____________ impurities whereas a n-type semiconductor material is doped with __________ impurities
A p-type semiconductor material is doped with electron-deficient (acceptor) impurities such as boron, whereas an n-type semiconductor material is doped with electron-rich (donor) impurities such as phosphorus.
A p-type semiconductor material is doped with electron-deficient (acceptor) impurities such as boron, whereas an n-type semiconductor material is doped with electron-rich (donor) impurities such as phosphorus.
See lessA 1mv of i/p gives an output of 1V, the voltage gain as such would be
The voltage gain (Av) can be calculated using the formula:[Av = frac{V_{out}}{V_{in}}]Given that ( V_{in} = 1 text{ mV} = 0.001 text{ V} ) and ( V_{out} = 1 text{ V} ), the calculation would be:[Av = frac{1 text{ V}}{0.001 text{ V}} = 1000]Therefore, the voltage gain is 1000.
The voltage gain (Av) can be calculated using the formula:
[
Av = frac{V_{out}}{V_{in}}
]
Given that ( V_{in} = 1 text{ mV} = 0.001 text{ V} ) and ( V_{out} = 1 text{ V} ), the calculation would be:
[
Av = frac{1 text{ V}}{0.001 text{ V}} = 1000
]
Therefore, the voltage gain is 1000.
See lessA 10 MVA, 11 kV alternator has positive, negative and zero sequence reactance of 25%, 40%, 10% respectively. The resistance that must be placed in the neutral such that fault current for a single phase to ground fault will not exceed the rated line current in ohms is __________
To calculate the resistance that must be placed in the neutral for a single-phase-to-ground fault such that the fault current does not exceed the rated line current, we can use the following relationship: 1. Rated Line Current (I_line):[I_{line} = frac{S_{rated}}{sqrt{3} times V_{LL}} = frac{10 , teRead more
To calculate the resistance that must be placed in the neutral for a single-phase-to-ground fault such that the fault current does not exceed the rated line current, we can use the following relationship:
1. Rated Line Current (I_line):
[
I_{line} = frac{S_{rated}}{sqrt{3} times V_{LL}} = frac{10 , text{MVA}}{sqrt{3} times 11 , text{kV}} = frac{10 times 10^6}{sqrt{3} times 11 times 10^3} approx 523.6 , text{A}
]
2. Fault Current (I_f) for a single-phase-to-ground fault:
– For a single-phase-to-ground fault in a generator, the fault current is determined primarily by the zero-sequence reactance (X0) since only one phase is involved.
– The relationship for the fault current is:
[
I_f = frac{V_{LL}}{Z_{fault}}
]
where (Z_{fault} = R + jX_0).
Given:
– (X_0 = 10% text{ of } Z_{base}).
– (Z_{base} = frac{V_{LL}^2}{S_{rated}} = frac{(11
See less