ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: golden on April 16, 2011, 12:40:21 pm
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I was thinking about this question (Chapter 4 Review from the Heinemann Textbook).
When the current goes through the circuit, it will first all go to Y and then through to X where it then passes through the rest or..?
Thanks!
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Ok the 20 ohm and 5 ohm resistors are in series together so add them up to 25ohm. That 25 ohm resistor is in parralel with the 10 and 5 ohm resistors. So that gives an effective resistance of 50/17 ohm = 2.94 ohm. The 10, 5 and 25 ohm resistors in parralel together are in series with the 10ohm resistor right at the top. The total resistance is therefore bout 12.94 ohm.
therefore the total current is 1.932 A.
now to find the voltage drop across the 10 ohm resistor.
10/12.94 x 25 =19.32 V
therefore the voltage drop across all the other resistors is 5.68 V
Since the 10 ohm(2nd one from the top) ,5 ohm and 25 ohm resistors are in parellel the voltage drop across them is identical.
so I = 5.68/25 = 0.227A
= 0.23 A
(current across the 20 and 5 ohm resistor at the bottom is the same since they are in series)
here is a very shit picture to show you what i mean heheh
[IMG]http://img64.imageshack.us/img64/4094/physsss.png[/img]
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I uploaded a picture with the current flow
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Ok the 20 ohm and 5 ohm resistors are in series together so add them up to 25ohm. That 25 ohm resistor is in parralel with the 10 and 5 ohm resistors. So that gives an effective resistance of 50/17 ohm = 2.94 ohm. The 10, 5 and 25 ohm resistors in parralel together are in series with the 10ohm resistor right at the top. The total resistance is therefore bout 12.94 ohm.
therefore the total current is 1.932 A.
now to find the voltage drop across the 10 ohm resistor.
10/12.94 x 25 =19.32 V
therefore the voltage drop across all the other resistors is 5.68 V
Since the 10 ohm(2nd one from the top) ,5 ohm and 25 ohm resistors are in parellel the voltage drop across them is identical.
so I = 5.68/25 = 0.227A
= 0.23 A
(current across the 20 and 5 ohm resistor at the bottom is the same since they are in series)
here is a very shit picture to show you what i mean heheh
[IMG]http://img64.imageshack.us/img64/4094/physsss.png[/img]
Wow, it's all clear now!
The picture is just as amazing as this:
http://www.youtube.com/watch?v=ULSG3pkjoT4&feature=fvwrel
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Wow, it's all clear now!
The picture is just as amazing as this:
http://www.youtube.com/watch?v=ULSG3pkjoT4&feature=fvwrel
Hahaha not quite as amazing :P
but yea with problems like these you just need to identify where the current is flowing and which bits are in series and in parallel.