my first coursework is due tomorrow, ive finished it all, but someone else in my lab has different answers, so i need help to see if my methods are sound:



Current at A is 5.9mA
Current at B is 3.9mA
Current at C is 2.7mA
Current at D is 0

Total Current is 12.6mA

i got this by:

13Volts / 220ohms = 5.9mA
13Volts / 330ohms = 3.9mA
13Volts / 470ohms = 2.7mA

Total Resistance = 1/.220 + 1/.330 + 1/.470 = 9.70

1/9.70 = Total Resistance = 103ohms

is my method sound or did i balls up somewhere?

its my first year at Uni and im a total electronics noob

Im sure one of the electronics guys can come here and sort this out properly.... but Id say you ballsed it up...

youre running on 13v all the time when its not ?

Id do the 8v with A and B as one circut

and then the 5v with just B and C in it ?



(i.e. at D youd have a pd between A and B, then a seperate pd between B and C ?)

its been a while since iv done this sort of thing but im pretty sure 13v wont be the V for all 3

this may sound dumb, but is youre diagram deffo right ?

(got a mate on msn taking a look )

QuoteDave says: I think you just count the batteris together and say its 13v
Alan says: really ?
Alan says: but the resistor at B in on the -ve line ?
Dave says: thats what I was wondering about, was wondering if hed drawn it wrong?
Dave says: if the batteris are facing opposite directions, treat it as two seperate circuits, then cancel out the middle bit
Dave says: so do A and B, then C and B
Dave says: then add the two B results together

cant garantee thats right... hes doing mechanical engeneering not electrical.... but its what I thought too....

The voltage at any point on that circuit will either be 5V, 8V or 3V, nowhere will have 13V because the orientation of the batteries mean the voltages will cancel, not add. Ive no idea what voltage would go where, as its a bit of an odd circuit. You sure one of the batteries shouldnt be the other way round?

edit, also you got the decimal point wrong in your calculations, theyd be 59, 39 and 27mA

whered you get 3v from ?

the only way I can see to do it is  run

8v through A and B
then
5vthrough C and B

work out the current flowing between A and B
the whats flowing between B and C

and add the 2 together

edit....

this


then this

thanks to everyones help, especially knighty on msn at this ungodly hour

ive got

A as 14.5mA (8V / 550ohms in series)
B as 20.7mA (A+C)
C as 6.25mA (5V / 800ohms in series)

yeah that Superposition Theorem gets pretty heavy after a few drinks

next time one of us should be sober  :?

As knighty says you have two batteries running in parallel, you cant just add them together. so you have to treat each as a separate entity. Then there is the resistance. If D is a switch then its a lot easier, you add 220ohms to 330 for the first circuit (8V) and then the 330 to 470 for the second (5v).

Dont expect me to do the math though as I am totally useless at it now. I would arse it up totally.