How to draw the output voltage waveform rectifier
1 answer:
Answer:
Half-wave rectifier converts an AC signal into a DC signal. It's called a half-wave because it only rectify the positive part of an AC signal.
AC Signal = An electrical signal that alternates between positive and negative voltage.
DC Signal = An electrical signal that only has positive voltage.
Rectify = A fancy word for converting something.
Adding a capacitor helps the positive part of the signal stay on longer. This work because the capacitor stores energy kinda like a battery. During the negative part of the AC signal, the energy stored in the capacitor will be drained and used, then the cycle repeats.
The load resistor is just there to prevent a short circuit from happening.
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Answer:
The temperature of the strip as it exits the furnace is 819.15 °C
Explanation:
The characteristic length of the strip is given by;
The Biot number is given as;
< 0.1, thus apply lumped system approximation to determine the constant time for the process;
The time for the heating process is given as;
Apply the lumped system approximation relation to determine the temperature of the strip as it exits the furnace;
Therefore, the temperature of the strip as it exits the furnace is 819.15 °C
Answer: The energy system related to your question is missing attached below is the energy system.
answer:
a) Work done = Net heat transfer
Q1 - Q2 + Q + W = 0
b) rate of work input ( W ) = 6.88 kW
Explanation:
Assuming CPair = 1.005 KJ/Kg/K
<u>Write the First law balance around the system and rate of work input to the system</u>
First law balance ( thermodynamics ) :
Work done = Net heat transfer
Q1 - Q2 + Q + W = 0 ---- ( 1 )
rate of work input into the system
W = Q2 - Q1 - Q -------- ( 2 )
where : Q2 = mCp T = 1.65 * 1.005 * 293 = 485.86 Kw
Q2 = mCp T = 1.65 * 1.005 * 308 = 510.74 Kw
Q = 18 Kw
Insert values into equation 2 above
W = 6.88 Kw
