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olga2289 [7]
3 years ago
11

Charging method .Constant current method​

Engineering
1 answer:
mina [271]3 years ago
4 0

Answer:

There are three common methods of charging a battery; constant voltage, constant current and a combination of constant voltage/constant current with or without a smart charging circuit.

Constant voltage allows the full current of the charger to flow into the battery until the power supply reaches its pre-set voltage.  The current will then taper down to a minimum value once that voltage level is reached.  The battery can be left connected to the charger until ready for use and will remain at that “float voltage”, trickle charging to compensate for normal battery self-discharge.

Constant current is a simple form of charging batteries, with the current level set at approximately 10% of the maximum battery rating.  Charge times are relatively long with the disadvantage that the battery may overheat if it is over-charged, leading to premature battery replacement.  This method is suitable for Ni-MH type of batteries.  The battery must be disconnected, or a timer function used once charged.

Constant voltage / constant current (CVCC) is a combination of the above two methods.  The charger limits the amount of current to a pre-set level until the battery reaches a pre-set voltage level.  The current then reduces as the battery becomes fully charged.  The lead acid battery uses the constant current constant voltage (CC/CV) charge method. A regulated current raises the terminal voltage until the upper charge voltage limit is reached, at which point the current drops due to saturation.

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sweet-ann [11.9K]

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social entrepreneurship

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3 years ago
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Question 1 : Replacement [42 Pts] Consider the following page reference string:
MakcuM [25]

Answer:

Optimal

Time 123456789101112

RS. ecbeagdcegda

F0. eeeeeeeee e e a

F1 c c c c c c c c c c c

F2 b b b g g g g g g g

F3 a a d d d d d d

Page fault? * * * * * * *

Total page fault:7

2. LRU

Time 1 2 3 4 5 6 7 8 9 10 11 12

RS e c b e a g d c e g d a

F0 e e e e e e e c c c c a

F1 c c c c g g g g g g g

F2 b b b b d d d d d d

F3 a a a a e e e e

Page fault? Y Y Y N Y Y Y Y Y N N Y Total page fault:9

3. LRU approximation algorithm: Second chance

Time 1 2 3 4 5 6 7 8 9 10 11 12

RS e c b e a g d c e g d a

F0 0,e 0,e 0,e 1,e 1,e 0,e 0,e 0,e 1,e 1,e 1,e 0,e

F1 0,c 0,c 0,c 0,c 0,g 0,g 0,g 0,g 1,g 1,g 0,g

F20,b0,b0,b0,b0,d0,d0,d0,d1,d0,dF30,a0,a0,a0,c0,c0,c0,c0,a

Page fault? YYYNYYYYNNNY

Total page fault: 8

4 0
3 years ago
Read 2 more answers
2. Similar to problem 1, assume your computer system has a 32-bit byte-addressable architecture where addresses and data are eac
andreev551 [17]

Question:

The question is not complete. The question to answer was not added. See below the possible question and the answer.

a. How many blocks are in the cache with this new arrangement?

b. Calculate the number of bits in each of the Tag, Index, and Offset fields of the memory address.

C. Using the values calculated in part b, what is the actual total size of the cache including data, tags, and valid bits?

Answer:

(a) Number of blocks =  512 blocks

(b) Tag is 18

(c)  Total size of the cache = 8388608 bytes

Explanation:

a .

block size = 32 bytes

cache size = 16384 bytes

No.of blocks = 16384 / 32

No.pf blocks = 512 blocks

b.

Total address size = 32 bits

Address bits = Tag + Line index +block offset

Block Size = 32 bytes.

So block size = 25 bytes.

Hence Offset is 5

No . of Cache blocks = 512 blocks = 29 blocks

Hence line offset is 9

We know that Address bits = Tag + Line index +block offset

So , 32 =tag+9+5

tag = 32-(9+5)

So Tag is 18

c.

Data bits = 32 bits

Tag=18 bits

Valid bit is 1 bit

so Total cache size = 25+218+20

                                  = 223

                                  =8388608 bytes

7 0
4 years ago
A joining process in which a filler metal is melted and distributed by capillary action between faying surfaces, the base metals
PolarNik [594]

Answer:

A soldering process

Explanation:

Given that ,The filler metal's melting point temperature is less than 450 ° C.

Usually, the brazing material has the liquid temperature of the melting point, the full melting point of the filler material approaching 450 degrees centigrade, while the filler material is less than 450 degrees centigrade in the case of soldering.

Therefore the answer is "A soldering process".

4 0
3 years ago
Calculate the LER for the rectangular wing from the previous question if the weight of the glider is 0.0500 Newton’s.
lapo4ka [179]

Answer:

0.2

Explanation:

Since the span and chord of the rectangular wing is missing, due to it being from the other question, permit me to improvise, or assume them. While you go ahead and substitute the ones from your question to it, as it's both basically the same method.

Let the span of the rectangular wing be 0.225 m

Let the chord of the rectangular wing be 0.045 m.

Then, the area of any rectangular chord is

A = chord * span

A = 0.045 * 0.225

A = 0.010 m²

And using the weight of the glider given to us from the question, we can find the LER for the wing.

LER = Area / weight.

LER = 0.010 / 0.05

LER = 0.2.

Therefore, using the values of the rectangular wing I adopted, and the weight of the glider given, we can see that the LER of the glider is 0.2

Please mark brainliest...

3 0
3 years ago
Read 2 more answers
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