1-10y - 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..
11ans is A ....... simple application of Kirchoff's laws
Actually ans is 0.03125A from P2 to P1 to be exact
1r u sure it is C mani......i checked .. it was given even in fiitjee solns...i even hav d solved paper.... it is A
3Ans b write kirchoff's law
5-12i110i2=0
i2=2+10i1
i1 and i2 are the currents thru the 5V and 2V batteries
1wats in name
dats why i hav written x * 2 in (1) it is current * int reseis off by the cell
1hey 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
1let 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
