cant_ride Posted June 3, 2005 Report Share Posted June 3, 2005 I can do pretty much everything in electronics apart from finding basic circuit parameters... dont have a clue why that is... and on that note i still have no clue how to find .... Ix and Vx in the following circuit ... help me out with some kind of walkthrough if you've got time. :) oli ( the dunce ) Quote Link to comment Share on other sites More sharing options...
Will Arnold Posted June 3, 2005 Report Share Posted June 3, 2005 (edited) i did things like that in the electric products lesson at gcse level and i cant remember for the life of me how to do it :"> i'll try and dig out some notes if i have time tonight :) ta Will Edited June 3, 2005 by WILL ARNOLD Quote Link to comment Share on other sites More sharing options...
CalopS Posted June 3, 2005 Report Share Posted June 3, 2005 V=R X I R= V/I I=V/R Quote Link to comment Share on other sites More sharing options...
Dr. Nick Riviera Posted June 3, 2005 Report Share Posted June 3, 2005 no v= i over r , unless im wrong about ohms law Quote Link to comment Share on other sites More sharing options...
Spikenipple Posted June 3, 2005 Report Share Posted June 3, 2005 no v= i over r , unless im wrong about ohms law ← Here's the basic Ohm's Law: Therefore they can all be placed in any position in the law, as long as the action between them is right (just like CalopS explained). Where's the input on that circuit by the way? :) Quote Link to comment Share on other sites More sharing options...
mink Posted June 3, 2005 Report Share Posted June 3, 2005 It's not quite a simple as ohm's law. You need to use Kirchoff's second law and Thevenin's theorem. But I've forgotten them... Quote Link to comment Share on other sites More sharing options...
Dr. Nick Riviera Posted June 3, 2005 Report Share Posted June 3, 2005 google is your friend google is still your friend Quote Link to comment Share on other sites More sharing options...
CalopS Posted June 3, 2005 Report Share Posted June 3, 2005 I cant work it out, im so going to pass my exams :) Im thinking I has to be 2A just like the other one too which would then make V = 48 because the total R = 24 But theres a part of me thinks I is differnt to the other one I really should know this :) Quote Link to comment Share on other sites More sharing options...
CalopS Posted June 3, 2005 Report Share Posted June 3, 2005 I cant see how those thoreys can help us get the answer, MUST REVISE MY ASS OFF! Quote Link to comment Share on other sites More sharing options...
cant_ride Posted June 3, 2005 Author Report Share Posted June 3, 2005 erm ... i think i might know ohm's and kirchoffs laws... plus thevenin, norton, superposition etc... like i said this is at a high level ... the hard part comes after ive worked this lot out.... but i can do the hard part if only i had Vx and Ix ... and u only need ohms and kirchoffs to work it out i know that much. oli Quote Link to comment Share on other sites More sharing options...
CalopS Posted June 3, 2005 Report Share Posted June 3, 2005 like i said this is at a high level glad it is :)" and u only need ohms and kirchoffs to work it out i know that much. But doesnt kirchoffs law just state that if you have a 10v supply or something and have say one resistor the voltage drop across that resistor must be 10 v???? I dont see how that can help us if we dont have the Voltage to begging with Quote Link to comment Share on other sites More sharing options...
cant_ride Posted June 3, 2005 Author Report Share Posted June 3, 2005 glad it is :)" But doesnt kirchoffs law just state that if you have a 10v supply or something and have say one resistor the voltage drop across that resistor must be 10 v???? I dont see how that can help us if we dont have the Voltage to begging with ← kirchoffs voltage law yeh ... but kirchoffs current law is what i need to use. :) ... well both of them actually I have the answers on a sheet I just cant seem to get to them oli Quote Link to comment Share on other sites More sharing options...
CalopS Posted June 3, 2005 Report Share Posted June 3, 2005 but kirchoffs current law is what i need to use whats that??> Quote Link to comment Share on other sites More sharing options...
Tartridge Posted June 3, 2005 Report Share Posted June 3, 2005 whats that??> ← google is your friend ← :) Quote Link to comment Share on other sites More sharing options...
cant_ride Posted June 3, 2005 Author Report Share Posted June 3, 2005 :) ← Ad... couldnt you have helped me instead of him... seeing as I started the topic... this really stressing me out GAAAAAAAAAAR!!! oli Quote Link to comment Share on other sites More sharing options...
Tartridge Posted June 3, 2005 Report Share Posted June 3, 2005 Haha, sorry Oil. http://www.google.com I know jack shit about electronics :) * *may not be true. Quote Link to comment Share on other sites More sharing options...
cant_ride Posted June 3, 2005 Author Report Share Posted June 3, 2005 Haha, sorry Oil. http://www.google.com I know jack shit about electronics :) * *may not be true. ← hahaha ... google teaches me the same as my books and my lecturer... which is the theory... not where im supposed to start ?? :) by the way my name's oli .. not oil .. hohoho :unsure: oli Quote Link to comment Share on other sites More sharing options...
tank_rider Posted June 3, 2005 Report Share Posted June 3, 2005 adding my 2p i'd say the resistance is 5.25ohms and voltage is 11.25V resistors in parallel are combined by 1/R+1/R=1/R resistors in series are added together R+R=R current (I) is constant throughout the circuit. Quote Link to comment Share on other sites More sharing options...
Tartridge Posted June 3, 2005 Report Share Posted June 3, 2005 by the way my name's oli .. not oil .. hohoho :) ← Oops :"> Sorry, hehe. I haven't really looked at it properly... but surely theres some way of splitting the circuit down into smaller bits, and then applying the theory to it? Just an idea... Quote Link to comment Share on other sites More sharing options...
cant_ride Posted June 3, 2005 Author Report Share Posted June 3, 2005 'fraid not tank rider ... and i need V and I ... not V and R come on theres gotta be a genius somewhere !!!! oli erm yeh splitting it is what i was trying to do but it still wont work out for me :) oli Quote Link to comment Share on other sites More sharing options...
tank_rider Posted June 3, 2005 Report Share Posted June 3, 2005 'fraid not tank rider ... and i need V and I ... not V and R come on theres gotta be a genius somewhere !!!! oli erm yeh splitting it is what i was trying to do but it still wont work out for me :) oli ← what sort of answers do you think it should be? Quote Link to comment Share on other sites More sharing options...
cant_ride Posted June 3, 2005 Author Report Share Posted June 3, 2005 Vx = 27 volts Ix = 0.6 Amps ?? oli edit.. guess i could try and work back from the answers ? Quote Link to comment Share on other sites More sharing options...
tank_rider Posted June 3, 2005 Report Share Posted June 3, 2005 right reading through my electronics book i have been being a tard :) ill have a bosh and see what i can come up with Quote Link to comment Share on other sites More sharing options...
cant_ride Posted June 3, 2005 Author Report Share Posted June 3, 2005 if you can do it i'll be permanently in love with you if i can work out how to resolve this kind of problem im gonna get 100% for the module exam no problem. just this bloody topic sucks!! oli Quote Link to comment Share on other sites More sharing options...
tank_rider Posted June 3, 2005 Report Share Posted June 3, 2005 well ive done it :unsure: get on your knees behatch :unsure: right first thing you need to find the current in the upper part of the circuit over the two parallel resistors (3 and 6). V through both resistors is the same. The voltage through the 3 ohm resistor is 6V, divide that by the 6 ohm resistor to get 1 amp flowing through it. Add those currents together to get 3A which is the current flowing through the top part. The total resistance from the top part is 9 ohms. The voltage this gives is 9x3=27V, just as you said it does :) Now onto the current in the lower part. The total resistance is 15 ohms. The total current passing through the lower part is 27/15=1.8 the 11 ohm resistor takes 11x1.8=19.8V from the 27. 27-19.8=7.2V left for the resistors in parallel. finally 7.2/12=0.6A :) you best be greatful (Y) Quote Link to comment Share on other sites More sharing options...
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