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![[Post New]](/templates/default/images/icon_minipost_new.gif) 9 Apr 2007 17:43:18 IST
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determine current in each branch in the attachemnt
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10 Jul 2008 18:39:03 IST
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current flowing through ef be I
through AB be a
through AD be b
than by symmetry current flowing through BC will be b and through DC a
and through BD take it i
now apply kirchoffs rule for loop
ADBA and BDCB and ABCFEA
YOU WILL GET THE ANSWER
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12 Jul 2008 08:47:01 IST
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let current to be I flowing through the circuit divide it as I1 I2 as u etered in branch keep dividing in every branch and apply Loop Rule u will get the three equations as
2 I1 + I3 - I2 = 0
2 I1 +5 I2 +2 I3 =2
I1 - 2 I2 - 4 I3 =0
solve it u will get the answer
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12 Jul 2008 21:09:18 IST
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A very simple way to go about this would be to employ the Star-Delta conversion.
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Will nip in at times to solve problems :)
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12 Jul 2008 21:16:02 IST
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yeah...we can just convert the Delta BCD into a Star - solve it accordingly and then the circuit will be simplified.
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Alcohol and Calculus -- DON'T MIX -- Please don't Drink and Derive!!!
A committee is a group of people who individually can do nothing but together can decide that nothing can be done.
MUNNA BHAI: Circuit, bole toh yeh Ford kya hai?
CIRCUIT: Bhai, gaadi hai.
MUNNA BHAI: Toh phir, yeh Oxford kya hai?
CIRCUIT: Bole toh, simple hai bhai, Ox mane Bael, Ford mane gaadi. Oxford bole toh Baelgaadi. |
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