In series connection of resistors, what happens to the current across each resistor?

Initially increases and then decreases

Magnitude of current across each resistor in parallel combination is

changes as per value of resistance

RESISTORS IN PARALLEL AND SERIES

Resistor in series and parallel

Resistors in Series Connection

T3-Week3-Resistors in AC Circuits - G12 ADVANCED

T3-Week 1-Resistors in AC Circuits

In this video we take a look at the 0:02 fetch to code 0:03 execute cycle including its effect on 0:06 the various registers we've previously 0:12 [Music] 0:14 discussed a computer is defined Definition 0:17 as an electronic device that takes an 0:20 input 0:22 processes data 0:25 and delivers output 0:29 in this simple example you can see we're 0:31 taking the input 5 0:35 we're multiplying it by 2 that's our 0:37 process 0:39 and we're outputting 10. 0:44 but this could be way more complex for 0:46 example of a game console 0:48 the input could be the buttons you press 0:50 on a controller 0:53 the processes would then be carried out 0:55 by the console itself 0:59 and the output would be some form of 1:01 update to a monitor 1:02 and sound out for a speaker possibly 1:04 vibration feedback through the 1:06 controller 1:10 to process data a computer follows a set 1:13 of instructions 1:14 known as a computer program 1:18 if we take the lid off a typical desktop 1:20 computer we can identify 1:22 two critical components the memory 1:26 that stores the program and the central 1:29 processing unit or processor 1:31 which is under this large fan and 1:33 carries out the instructions 1:37 a computer carries out its function by 1:40 fetching 1:41 instructions decoding them and then 1:43 executing them 1:44 in a continuous repetitive cycle 1:46 billions of times a second 1:48 let's look at each of these stages in a 1:50 little more detail Fetch 1:53 so let's start with the fetch stage the 1:55 very first thing that happens 1:57 is the program counter is checked as it 2:00 holds the address 2:01 of the next instruction to be executed 2:07 the address stored is then copied into 2:09 the memory address register 2:14 the address is then sent along the 2:16 address bus to main memory 2:18 where it waits to receive a signal from 2:21 the control 2:22 bus so it knows what to do 2:27 as we want to read the data that's 2:29 stored in memory address 2:30 0 0 0 0 the control unit sends 2:34 a read signal along the control bus to 2:36 main memory 2:41 now main memory knows the data needs to 2:44 be read 2:45 the content stored in memory address 000 2:49 can be sent along the data bus to the 2:51 memory data register 2:56 now as we're currently in the process of 2:58 fetching an instruction 3:00 the data received by the memory data 3:03 register gets copied 3:04 into the current instruction register 3:11 the instruction effectively has now been 3:14 fetched from memory 3:16 just before we proceed to the decode 3:18 phase we now 3:19 increment the program counter so that 3:22 the address it contains 3:24 points to the address of the next 3:26 instruction which will need to be 3:30 executed 3:32 the instruction now being held in the 3:33 current instruction register 3:35 is ready to be decoded 3:39 now as we mentioned in the previous 3:41 video the instruction is made up of two 3:43 parts 3:44 we have the op code that's what it is we 3:47 need to do 3:50 and we have the operand what are we 3:53 going to do it to 3:55 now the operand could contain the actual 3:57 data 3:58 or indeed it could contain an address of 4:01 where the data is to be found 4:06 by decoding this instruction we can see 4:08 the operation we need 4:10 is a load operation so we need to load 4:14 the contents of memory location0101 4:18 into the cpus accumulator 4:25 in the exam a simple model will be used 4:27 to describe the 4:29 structure of any given instruction 4:32 you're not going to be expected to 4:34 define how an opcode is made up 4:36 but simply to interpret opcodes in the 4:39 given context of an exam 4:40 question in the example here 4:44 you can see there's a total of 16 4:46 different opcodes available 4:48 and this is because we're using four 4:50 bits for our representation 4:56 so now we've fetched the instruction and 4:59 we've decoded it so we know what we need 5:00 to do 5:01 we're finally ready to execute it 5:05 so we now send address 0101 5:08 to the memory dress register 5:13 now we're in the memory address register 5:15 we can finally send the address 5:18 down the address bus to main memory 5:24 this time we want to read the data 5:26 that's stored in memory 5:28 and so the control unit again sends a 5:30 read signal along the control bus 5:36 so main memories now receive an address 5:38 and a read signal 5:40 so the content stored at memory location 5:43 0101 5:44 can now be sent along the data bus back 5:46 to the cpu 5:47 and into the memory data register 5:54 finally the contents of the memory data 5:56 register are copied to the accumulator 5:59 and this is one of a number of general 6:00 purpose registers found in the cpu 6:04 this first instruction is now complete Branching 6:11 so what does this program actually do 6:14 you should be able to work it through 6:16 carefully and figure it out 6:19 we're now pointing instructions zero 6:21 zero zero one in the program counter 6:23 and we're ready to fetch the second 6:25 instruction 6:27 at the end of this video we're gonna 6:29 provide you with the answer 6:34 so let's talk a second about programs 6:37 that branch 6:40 on the left here we have a very simple 6:42 piece of pseudo code 6:44 line zero says first execute this line 6:46 of code 6:47 line 1 now execute this line and then 6:50 line 2 says 6:52 if the age is greater than 18 then 6:56 we're going to execute lines 3 and 4 6:58 otherwise 6:59 we're going to execute lines six and 7:02 seven 7:03 so this program doesn't necessarily 7:05 follow strictly in sequence from line 7:07 zero through to seven there's a chance 7:10 here the program may branch and jump 7:14 around 7:16 so we're going to pretend that this 7:17 program has been loaded into memory 7:20 each line of code on the left here has 7:23 ended up 7:24 as a location in memory now this is not 7:27 strictly how this would happen in this 7:28 one-to-one way 7:29 but for the purpose of example it's 7:31 absolutely fine 7:35 so the program counter starts by 7:37 pointing to memory address zero 7:39 and we fetch the first instruction 7:41 decode it and execute it 7:44 it then updates and tells us the next 7:47 instruction 7:48 is zero zero zero one because remember 7:50 the program counter is being incremented 7:52 so we fetch it decode it and we execute 7:55 line one of our program 7:59 we then fetch line two which in binary 8:01 is one 8:02 zero 8:06 now at this point depending on what 8:10 happens during the execution 8:11 of line two the program may be required 8:15 to fetch line three from memory or 8:18 line five from memory 8:25 so let's look at how this actually works 8:27 because we've said the program counter 8:28 simply gets incremented 8:31 well in the current instruction register 8:33 we have an instruction with the op code 8:36 0 1 1 0. 8:41 now when we look this up in the decode 8:43 unit we discover that this 8:45 code means branch always 8:51 this replaces the value held in the 8:54 program counter 8:56 with the contents of the operand that's 8:58 the second part of the instruction 9:01 from the current instruction register so 9:03 this case 9:04 one zero zero one 9:09 now when the next fetch cycle begins the 9:12 program counter is obviously checked 9:14 and as its contents have been previously 9:16 updated to a new memory location 9:19 and not simply incremented the program 9:22 effectively is able to jump 9:24 around memory 9:28 so having watched this video you should 9:30 be able to answer the following key 9:32 question 9:33 how does a cpu work 9:39 okay so let's um answer the question we 9:41 posed 9:42 earlier what did that program actually 9:48 do 9:50 so this is the first fetch to code 9:53 execute cycle 9:55 and this is the one that we ran through 9:57 in detail earlier 9:58 it effectively loaded the contents of 10:01 the memory 10:02 stored at location location0101 10:05 into the accumulator in other words 10:08 the dna number 3 is moved 10:11 from memory into the cpu 10:18 we then proceed onto the second fetch 10:20 decode execute cycle 10:23 now this one adds the contents of memory 10:27 located at 0 1 1 0 10:30 to the current contents of the 10:32 accumulator 10:34 so in other words the dna number one 10:38 because that's what's stored at address 10:40 zero one one zero 10:43 is added to the number three that was in 10:45 the accumulator 10:46 the results are stored back over the 10:48 accumulator 10:49 so effectively we've done three plus one 10:53 equals four 10:58 the third fetch to code execute cycle 11:00 stores the contents which are in the 11:02 accumulator 11:03 into memory location zero one one one 11:07 and that's because the op code the first 11:09 part of this current instruction 11:10 zero zero one one is the command to 11:13 store when we look it up in the decoder 11:15 unit 11:16 so in other words the result of the 11:17 previous calculation three plus one 11:19 equals four 11:20 is now written back into main memory 11:28 the fourth fetch decode execute cycle 11:30 outputs the contents of the accumulator 11:33 remember they were copied into main 11:34 memory but they're still held in the 11:35 accumulator 11:37 so in this simple abstraction the number 11:40 four is now 11:41 output to the user so they can see the 11:43 result of the calculation 11:49 the fifth and final fetch code execute 11:51 cycle 11:52 brings a halt to the current program 11:58 so this very simple program which has 12:01 five 12:02 fetch decode execute cycles has 12:04 performed the calculation 12:06 three plus one is then stored the result 12:09 in main memory 12:10 and displayed the result four to the 12:12 user 12:13 and in a high-level language this may 12:15 look something very similar to the 12:17 following two lines of code 12:20 sum variable equals num1 plus num2 12:24 print sum to the user 12:27 so you can start to get an appreciation 12:29 here of how the high level code you 12:32 write actually ends up being fetched 12:34 decoded 12:35 and executed inside a processor 12:38 of course your processor is doing 12:40 billions and billions of these 12:42 operations a second 12:43 which when you think about it is really 12:45 very impressive 12:52 [Music] 13:03 you. make 10 questions for a standerd of a level

Integumentary system: skin and its accessory structures such as hair, sebaceous glands, sweat glands and nails. Antigen presenting cells (APCs)- a heterogeneous group of immune cells that mediate the cellular immune response by processing and presenting antigens for recognition by certain lymphocytes such as T cells. Classical APCs include dendritic cells, macrophages, Langerhans cells (macrophages of the skin that can migrate to the lymph vessels) and B cells. Hemidesmosomes- multiprotein complexes that facilitate the stable adhesion of basal epithelial cells to the underlying basement membrane. Type I collagen- fibrils form fibers which can form bundles; resists tension (most abundant; in tendons, bones, dermis) Type VII collagen- collagen type essential in strengthening and stabilizing the skin.

1. Da cosa si ricavano le materie plastiche? a) Sabbia, calce e argilla b) Argilla e acqua c) Petrolio e derivati d) Gesso e ferro 2. Qual è uno svantaggio delle plastiche tradizionali? a) Costano troppo b) Si degradano molto lentamente c) Sono conduttrici di elettricità d) Contengono ossigeno e idrogeno 3. Le materie plastiche sono utili perché: a) Sono sempre biodegradabili b) Assorbono acqua e luce c) Resistono bene solo alle alte temperature d) Sono leggere, modellabili e resistenti 4. Quali sono i principali materiali artificiali da costruzione? a) Rame, ferro, zinco e piombo b) Laterizi, ceramica e vetro c) Sabbia, ghiaia e marmo d) Legno, carbone e calcare 5. Come si classificano i laterizi in base all’uso? a) Opachi e trasparenti b) Per strutture, coperture, pavimenti e decorazioni c) Solidi e liquidi d) Per interni ed esterni 6. I laterizi sono ottenuti da: a) Argilla cotta b) Marmo frantumato c) Vetro fuso d) Plastica riciclata 7. I materiali leganti vengono usati per: a) Rivestire pareti b) Unire materiali da costruzione c) Colorare la ceramica d) Lucidare superfici 8. Il cemento si ottiene da: a) Cottura di Sabbia e acqua b) Cottura di Calce viva e acqua c) Cottura di calcare e argilla d) Fusione del ferro 9. Quale materiale è usato per il cemento armato? a) Acciaio b) Mattoni pieni c) Fibra di vetro d) Rame lucido 10. Il vetro si produce principalmente dalla: a) Calce e argilla b) Silice e ossidi c) Resina e olio d) Gesso e caolino 11. Il vetro pyrex è utilizzato per: a) Tetti e coperture b) Contenitori resistenti al calore c) Oggetti decorativi in legno d) Pavimentazioni industriali 12. Il metodo "float glass" serve per produrre: a) Tubi in PVC b) Lastre di vetro c) Mattoni refrattari d) Bottiglie in plastica 13. Qual è una proprietà del vetro? a) È biodegradabile b) È un isolante elettrico c) Assorbe umidità d) Conduce il calore facilmente 14. Un vantaggio dell’uso delle plastiche è che: a) Sono facilmente modellabili b) Si sciolgono con l’acqua c) Sono trasparenti come il vetro d) Durano poco 15. Le rocce metamorfiche sono: a) Rocce trasformate da calore e pressione b) Completamente artificiali c) Usate solo per decorare d) Composte da sabbia e gesso 16. Il calcestruzzo è formato da: a) Solo sabbia e cemento b) Cemento, sabbia, acqua e ghiaia c) Argilla e vetro d) Cemento e legno

A. The continental plate remains above, while the oceanic plate subducts. The older and denser oceanic plate subducts beneath the younger plate. A subduction is the process by which an oceanic plate slide under a less dense plate, can be a continental or another oceanic plate. In this process, the plates involved are oceanic plate and continental plate. Oceanic Plate is thinner plate, Dense,generally, slides under into the mantle; and especially when it is older than the other oceanic plate. They may slide over given that it is the younger oceanic plate in the oceanic-oceanic subduction. In case of oceanic-continental subduction, even younger oceanic plate can never slide over it. Continental Plate is thicker plate, less dense and slides over and experiences compression and volcanic activity. Another concept is about buoyancy. Consider the Earth’s mantle as a giant swimming pool. Floating on top of it are the Earth's tectonic plates—some thin and dense, like the oceanic crust, and others thick and less dense, like the continental crust. Imagine the thin oceanic plate and the thick continental crusts are like tennis ball and soccer ball, respectively. When placed in water, the tennis ball sinks at the bottom and easily subducted as it is smaller and denser. Contrary, the soccer ball is larger and more buoyant, thus, resists subduction and tends to stay afloat. Therefore, during subduction, the thin and dense oceanic crust or an older oceanic crust slide under another plate due to its low buoyancy. Consequently, the thick and denser continental plate or younger oceanic plate slides over because of its high buoyancy. In the subduction zone, there are two landforms that are formed in the process, namely, trench and volcanic arcs. Trenches are deep valleys formed at the edges of the colliding plates, where an oceanic plate bends downward and starts to subduct another plate. It looks like a long, narrow depression where marks the zone where subduction begins. The other landform is the volcanic arc, a chain of volcanoes that formed on the overriding plates where water and sediments from the sinking slab cause the mantle wedge above it to melt, making the magma rise to the surface and forming the volcanoes. a. The formation of trench When these two plates collide, the oceanic plate is subducted and pulled into the mantle. The edges of the plates create a deep valley which we call trench. b. Formation of volcanic arcs When a denser oceanic plate collides with a less dense continental plate, oceanic plate is subducted and pulled into the mantle. As it reaches the mantle, the plate is subjected to extremely high pressure and temperature. This causes the trapped water and air in the plate to be released. The plate eventually melted back as magma. The formed magma rises to the surface. This gives rise to the formation of volcanic arcs. The same is true between the collision of two oceanic crusts as the older crust is subducted over the younger crust. However, the collision of two continental crusts will not result in the formation of trenches.

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