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Old 16-05-12, 08:45 PM   #1
davepreston
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Default electronics (maths) question



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

Last edited by davepreston; 16-05-12 at 08:47 PM.
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Old 16-05-12, 08:50 PM   #2
dyzio
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Default Re: electronics (maths) question

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

Last edited by dyzio; 16-05-12 at 08:53 PM.
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Old 16-05-12, 08:53 PM   #3
flymo
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Default Re: electronics (maths) question

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.

Last edited by flymo; 16-05-12 at 08:56 PM.
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Old 16-05-12, 08:54 PM   #4
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Default Re: electronics (maths) question

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Old 16-05-12, 09:17 PM   #5
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Default Re: electronics (maths) question

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.
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Old 16-05-12, 09:20 PM   #6
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Default Re: electronics (maths) question

thanks lads
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Old 16-05-12, 09:21 PM   #7
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Default Re: electronics (maths) question

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
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Old 16-05-12, 09:56 PM   #8
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Default Re: electronics (maths) question

Has nobody thought to ask Dave why he wants to know this? It all seems a little fishy to me
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Old 16-05-12, 09:56 PM   #9
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Default Re: electronics (maths) question

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.

Last edited by tigersaw; 16-05-12 at 10:00 PM. Reason: wrote series instead of parallel
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Old 16-05-12, 10:02 PM   #10
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Default Re: electronics (maths) question

Quote:
Originally Posted by theboatman View Post
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 !
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