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3 years ago

5

An intelligence signal is amplified by a 65% efficient amplifier before being combined with a 250W carrier to generate an AM sig

nal. If it is desired to operate at 50% modulation, what must be the dc input power to the final intelligence signal amplifier

1 answer:

Aleks04 [339]3 years ago

3 0

Answer:

"192.3 watt" is the right answer.

Explanation:

Given:

Efficient amplifier,

= 65%

or,

= 0.65

Power,

$P_c=250 \ watt$

As we know,

⇒ $P_t=P_c(1+\frac{\mu^2}{2} )$

By putting the values, we get

$=P_c(1+\frac{1}{2} )$

$=1.5 \ P_c$

Now,

⇒ $P_i=(P_t-P_c)$

$=1.5 \ P_c-P_c$

$=\frac{P_c}{2}$

DC input (0.65) will be equal to " $(\frac{P_c}{2} )$ ".

hence,

The DC input power will be:

= $\frac{250}{2}\times \frac{1}{0.65}$

= $\frac{125}{0.65}$

= $192.3 \ watt$

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Answer:

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The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"

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$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

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$\tau_{max} = \frac{4}{3} \cdot \frac{V}{A}$

Where:

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$A$ - Cross section area, measured in square meters.

As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:

$V = \frac{P}{5}$

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$A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}$

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The diameter is now cleared and computed:

$D = \sqrt{\frac{4}{\pi}\cdot A}$

$D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})$

$D = 0.0457\,m$

$D = 45.7\,mm$

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