The 12V looks to be the drop across R2
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Discuss Trainee questions to answer in the Electrical Course Trainees Only area at ElectriciansForums.net
Right then, i was thinking of the circuit as being in series which is all ive seen so far so i thought the current was common but since its in parallel (which i have been reading up on all yesterday!) its the voltage that is common so i can use the 12v drop across R2 to give me the current at R1... so 12v / 4ohms = 3amps through R1
R1 + R2 = 5.4amps at R3
So the voltage drop across R3 is 5.4amps x 6ohms which is 32.4volts (hopefully!)
Correct stu !
Well done mate !
Sorry if the drawing was not clear enough, thats just how the tutor at college had drawn it on the board so I copied it
Have you got a question to put up for the rest of us ?
Cheers
.... see if I can work it out ( doubt it though lol )
Hmm.. we have only just started looking at this the other night so dont think ill be able to get this one just yet, will have a read of my notes though and see if I can work it out ( doubt it though lol )
I found that one a bit tricky when we did these at college, if you struggle with it i can post up one of the example equations that we were given before we started the exercise as these were a real help when we were left to get one with questions.
Stu
great start, NOT !! lol
Hi pennychew, we've just started this at college, so I'll give it a go! Is the answer 10mm2 cross sectional area? Cheers. Jim
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