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3 years ago

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1- A square-wave inverter has a dc source of 96 V and an output frequency of 60 Hz. The load is a series RL load with R = 5 Ohm

and L = 100 mH. Determine (a) Minimum and maximum Current (b) an expression for load current, (c) Amplitudes of the Fourier series terms (1,3,5,7,9) for the square wave load voltage (d) Amplitudes of the Fourier series terms (1,3,5,7,9) for the load current.

1 answer:

Helen [10]3 years ago

3 0

Answer:

D

Explanation:A circuit is equivalent to another when the output voltage is equal to each other. This is so because a circuit may have a higher voltage source and lesser load or a lower voltage source and higher load,so we can only say they are equal If the output is similar

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(a) If 5 x 10^17 phosphorus atoms per cm3 are add to silicon as a substitutional impurity, determine the percentage of silicon a

Y_Kistochka [10]

Answer:

The percentage of silicon atoms per unit volume that are displaced in the single crystal lattice = 0.001 %

The percentage of silicon atoms per unit volume that are displaced in the single crystal lattice with boron atoms = 0.4 × $10^{-5}$ %

Explanation:

No. of phosphorus atoms = 5 × $10^{17} \ cm^{-3}$

The volume occupied by a single Si atom

$V_{si} = \frac{a^{3} }{8}$

$V_{si} = \frac{5.43^{3}(10^{-8} )^{3} }{8}$

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$n_{si} = \frac{1}{V_{si} }$

$n_{si}$ = 5 × $10^{22}$ $\frac{atoms}{cm^{3} }$

$PCT = \frac{N_p}{N_{si}} 100$

Put the values in above equation we get

$PCT = \frac{5 (10^{17} )}{5 (10^{22}) } 100$

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(b).

No. of boron atoms = 2 × $10^{15} \ cm^{-3}$

The volume occupied by a single Si atom

$V_{si} = \frac{a^{3} }{8}$

$V_{si} = \frac{5.43^{3}(10^{-8} )^{3} }{8}$

$V_{si} =$ 2 × $10^{-23}$ $\frac{cm^{3} }{atom}$

$n_{si} = \frac{1}{V_{si} }$

$n_{si}$ = 5 × $10^{22}$ $\frac{atoms}{cm^{3} }$

$PCT = \frac{N_p}{N_{si}} 100$

Put the values in above equation we get

$PCT = \frac{2 (10^{15} )}{5 (10^{22}) } 100$

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These are the percentage of silicon atoms per unit volume that are displaced in the single crystal lattice.

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Answer:

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Explanation:

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$\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}+\dfrac{d^2T}{dz^2}=\dfrac{1}{\alpha }\dfrac{dT}{dt}$

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$\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}=\dfrac{1}{\alpha }\dfrac{dT}{dt}$

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