Electricity

-10y - 2x + 5 =0
==> 2x + 10y =5------1

sry... in first line it is x and not 'd'

In the square loop in the left hand side....

x * 2 (int resis) - 10 * y(as assumed as current in the branch P2 P1) = 5(battery's voltage

I am gettin as A .... i may b wrong... if yes then correct it..

ans is A ....... simple application of Kirchoff's laws

Actually ans is 0.03125A from P₂ to P₁ to be exact

yup.. ans ia A...

ans IS A !

its last years's q if am not wrong....

r u sure it is C mani......i checked .. it was given even in fiitjee solns...i even hav d solved paper.... it is A

MSP read #9

Ans b write kirchoff's law

5-12i₁10i₂=0

i₂=2+10i₁

i₁ and i₂ are the currents thru the 5V and 2V batteries

Bye the way answer given is C

-10y - 2x + 5 =0
==> 2x + 10y =5------1

explain this thing!!!

hey bt it has internal ressitnce na it will mattr

so u shud take d terminal voltage n nt d correct emf of d cell

this is wat i think

let d be the current flowin from 5V battery and y in the branch having 10 ohms......ie x-y is the current in the branch having 2V battery

Appling KVL

AP2P2CA and P2BDP1P2

-10y - 2x + 5 =0
==> 2x + 10y =5------1
and
+2 - 1(x-y) +10 y=0
+x-11y = 2
=> 2x -22y = 4-----------2

1 - 2
gives y =1/32

=0.03 amp from p2 to p1

koi isko to bata do
pleaseeeeeeee