electronics (maths) question

[davepreston]

http://natscareers.co.uk/images/network_resistance_graphic.jpg

whats the resistance in the network and how would you work it out

[dyzio]

Calculate:
1) R4 and R5 in parallel (R4XR5)/(R4+R5)
2) add R3 + answer 1) - series connection,
3) R2 and answer 2) in parallel
4) Add R1 + R6 + answer 3

4) = RTotal

[flymo]

250 ohms in total

the formulas above are good, simple way though when the numbers are easy enough. Two resistors of equal value in parallel - halves the resistance. So R4 and R5 can be replaced by 50ohms, then add R3 making 100 ohms. Same again between R2 and (R3+R4+R5 combo) is 50 ohms.

Then add the resistances in series = 100 + 50 + 100 = 250 ohms.

[dizzyblonde]

42:scratch:

[yorkie_chris]

All you should look first in any resistor network is how to split it into blocks which can be calculated with formulas you have to simplify them.

[davepreston]

thanks lads

[Specialone]

Im leaving this thread quick, my brain is full, in order for me to take these calcs on board, i'd have to push something out :)

[theboatman]

Has nobody thought to ask Dave why he wants to know this? It all seems a little fishy to me

[tigersaw]

AS YC says, break it down and the answer will fall out.

The guy who stares at it for an uncomfortable minute then says 250 ohms will be asked how he worked it out, (notice the question asks 'whats the resistance in the network and how would you work it out ') so the best approach would be to say along the lines of:
The resistance of the network is going to be the sum of the resistors in series going round the loop. So to simplify it, R4 and R5 are two 100 ohm resistors in parallel which makes 50 ohms*, which are in series with R3 which is another 50 ohms so that is 100 ohms. That network is itself in parallel with R2, so we have another parallel network of two 100 resistors, making 50 ohms. Add that to R1 and R6 and the total is 250 ohms.

The next thing you may be asked is ' notice there is a 12v battery in the circuit, can you tell me how you would work out the current flowing in the circuit?'

The answer would be to mention ohms law, saying that V=IR (voltage is equal to current times resistance), so that the current would be 12 (volts) divided by 250 (ohms) - which is about 1/25th of an amp.

Brownie points have been scored by knowing the total resistance is the sum of the resistors in series, knowing how to break up the diagram to work out the result, and if asked further, knowing ohms law.

Knowing how to work it out but getting it wrong will still score brownie points, just saying you dont know or blurting out a guess (even if right) wont.

* Dyzio is right with the formula R4 and R5 in parallel (R4XR5)/(R4+R5) - but if both resistors are the same value, as is often the case in these simple tests, then its just half. Similarily if three resistors are in parallel all the same value then its a third. So three 100 ohm resistors in parallel equate to a single resistor of 33 (and a third) ohms.

[Teejayexc]

Has nobody thought to ask Dave why he wants to know this? It all seems a little fishy to me

He's training to be a sparky?


God have mercy on our souls !

[Bibio]

please tell me your not applying for the engineers job at kirkcaldy vic

[yorkie_chris]

Here's one for any of you clever people who might know owt about batteries.

2 ways of wiring up some (NiMH, if it makes any odds...) cells. Both would give the same total voltage, and same total current. Any difference? Or difference in charging behaviour?

http://desmond.imageshack.us/Himg98/scaled.php?server=98&filename=cells.jpg&res=landing

[Jackie_Black]

is the left one series and the right one series parallel?

[yorkie_chris]

I suppose 1 is 2S2P and one 2P2S, however I dunno if it will make any difference to cell balancing.

Basically I am building a dive light which should be sufficient to cook fish without the inconvenience of removing them from the sea first :-P

[Jackie_Black]

Sweet!!

I think the right one charges at twice the speed if wired right.

[tigersaw]

The second diagram is the same as the first just with the anodes of the lower cells joined. There is more chance of a failed cell discharging adjacent cells, or thermal runaway.