Electricity and magnetism - grounded circuit

33

Abhishek Priyam ·2008-12-25 08:24:28

Ok..

Pot at grounded point is zero...

Potential at A is clearly +12 ...(is it clear?)

pot at D is -6... (clear?)

so pot diff across resistances is 18..

so i=3

so potential diff accross 1 ohm is 3 (3*1) (.....I*R) so pot of B=12-3=9

pot diff across 2 ohm is (3*2) so pot at C=9-6=3

pot diff across 3 ohm is (3*3) so pot at D =3-9 = -6 (which was clear above and is proof that there is no calaculation mistake above

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33

Abhishek Priyam ·2008-12-25 08:47:22

Sky: heheheheheheh
hehehhehehehhehe
hehehehhehehehhehehehhe
tell him wid asian paints

me: :)

Sky: bazaar se lake dal diya!

me: :D

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1

skygirl ·2008-12-25 08:45:41

bazaar se asian paints lakey dal diya :P

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1

Akshay Pamnani ·2008-12-25 08:32:24

How did u colour your post??

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1

Akshay Pamnani ·2008-12-25 08:30:51

How did u colour your post??

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3

msp ·2008-12-25 08:30:23

thanks for ur soln and thanx i get cleared in my doubt

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1357

Manish Shankar ·2008-12-25 08:29:14

Yeah I'm doing

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1

Akshay Pamnani ·2008-12-25 08:26:02

see the point earthed has 0 potential,so lower terminal of upper battery has 0 potential
Now upper terminal of lower battery will be at 0 potential
and the upper terminal of upper battery at 12V,and lower terminal of lower battery at -6 V
also equivalent of batteries=18V
18=6I
I=3A
Remaining can be easily done