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ruby92:

--- Quote from: astarmathsandphysics on March 27, 2010, 02:21:08 pm ---V2+V3=3
I2=2I3 whe I2 is current in 2 ohm resistor etc  by symmetry
2I2+3I2=3 so I2=3/5A IA=IB=3/5A

--- End quote ---
this is the answer to the first question?

astarmathsandphysics:
Net voltage around left loop is 1V
V=IR so 1=6I so I=1/6A
similarly I2=1/5 and I3=1/5+1/6 sine the currents are opposed

ruby92:
this is what i did
in the circuit a third current -Ia-Ib is there
for the first loop going in an anti clockwise direction
2Ia+3(-Ia-Ib)-2=0
2Ia-3Ia-3Ib-2=0.
-1Ia-3Ib-2=0.....(1)
in the second loop.
going anticlockwise
-2Ib-3(-Ia-Ib)+2=0
-2Ib+3Ia+3Ib+2=0
1Ib+3Ia+2=0....(2)

solving simulatneously.....
Ib as -o.5
and Ia by substitution as -0.5

what exactly have i done wrong?

Phosu:
its weird that i got this rite in the exam, but i cant solve it again,
haha heres how i did it.

Ia + Ib = Ic , heres the basic mistake which you did, which caused the whole thing to go rong.

Ia be a, and Ib be b

2-2a-3Ic=0
2-2b-3Ic=0

any 2 above can be used as equation 1

2-2a-3Ic=2-2b-3Ic

3Ic and 2 will get cancelled
you get an equation -2a+2b=0, this will be used as equation 2

now simultaneously solve
i will use loop a, and split Ic current

2-2a-3(a+b)=0
2-2a-3a+3b=0

1)2-5a+3b=0
2)-2a+2b=0

and voila! :D

ruby92:
whts d final answer???
and wht have i done wrong??
Reply fast

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