Answer:
Explanation:
If the museum is running short of funds, and you decide to increase revenue. An increase or decrease in the price of admission into the museum depends on the following:
1. If demand for admission into the museum is elastic there are two possible outcomes
a. An increase in the price of admission leads to a decrease in the quantity demand of admission into the museum
b. A decrease in price of admission into the museum leads to an increase in the quantity demand of admission into the museum.
This follows the law of demand which states that "the higher the price, the lower the quantity demanded and the lower the price, the higher the quantity demanded".
2. If the demand for admission into the museum is inelastic, then an increase in price will lead to an increase in revenue of the museum.
Therefore, before the curator increase the price of admission into the museum, he should first determine the price elasticity of demand of the museum.
Answer:
The frequency that the sampling system will generate in its output is 70 Hz
Explanation:
Given;
F = 190 Hz
Fs = 120 Hz
Output Frequency = F - nFs
When n = 1
Output Frequency = 190 - 120 = 70 Hz
Therefore, if a system samples a sinusoid of frequency 190 Hz at a rate of 120 Hz and writes the sampled signal to its output without further modification, the frequency that the sampling system will generate in its output is 70 Hz
Answer:
(a) 16.27 Vpk
(b) 48.7%
Explanation:
The transformer is assumed to be an ideal 10:1 voltage divider with no internal impedance. The diode is assumed to be modeled in the forward direction by a perfect 0.7 V voltage drop with no internal impedance. That means the frequency of the supply voltage is irrelevant.
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<h3>(a)</h3>
The peak voltage will be 0.7 V less than the transformer secondary peak voltage:
((120 V)√2)/10 -0.7 V ≈ 16.27 V
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<h3>(b)</h3>
The fraction of the amplitude for which the diode is non-conducting is ...
0.7/(12√2) ≈ 0.041248
The period of conduction is symmetrical about the peak of the waveform, so it is convenient to use the arccos function to find the (half) conduction angle:
arccos(0.041248) ≈ 87.64°
As a fraction of half the cycle, this is ...
conduction fraction ≈ 87.64°/180° ≈ 48.7%