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%
Answer:the entropy change for the process is 8.736kj/k
Explanation: the first step is to get the entropy change for iron, copper and that of lake as well. To get further explanation please see attached a solved copy of the problem
Answer:500,551.02
Explanation:
Given
Initial enthaly of pump \left ( h_1\right )=500KJ/kg
Final enthaly of pump \left ( h_2\right )=550KJ/kg
Final enthaly of pump when efficiency is 100%=
Now pump efficiency is 98%
=
0.98=
therefore initial and final enthalpy of pump for 100 % efficiency
initial=500KJ/kg
Final=551.02KJ/kg
The general rule is the larger the battery the more capacity/energy storage.