In the circuit shown in fig. calculate the following:
a. Potential differece between points a and b when switch S is open.
b.Current through S in the circuit

Explanation

i. Here `I_(1) = (36 - 0)/(9) = 4A`
Also `I_(2) = (36 - 0)/(9) =4A`
Also, `36 - V_(alpha) = I_1xx6 = 6I_1`
= 6xx4 =24V
`:. V_(alpha) = 36 - 24 = 12V`
`36 -V_(b) =3xx4 =12C`
and `V_(b) = 36 - 12 =24V`
`:. V_(alpha) -V_(b) =12 24 =-12V`

ii. in loop ABCIHA.
`-6(I - I_(1)) + 3I_(1) = 0 or -6I + 6I_(1) +3I_(1) = 0`
or `9I_(1) = 6I`
` :. I = (9)/(6)I_(1) = (3)/(2)I_(1)`

in loop CDEJIC,
`-3(I - I_(1) +I_(2)) + 6(I_(1) - i_(2)) =0`
or `-3I +3I _(1) - 3I_(2) +6I_(1) - 6I_(2) = 0`
or `-3I + 9I_(1) - 9I_(2) = 0` (ii)
in loop EFGHIJE,
`36 -3I_(1) - 6(i_(1) - I_(2)) = 0`
or `9I_(1) - 6I_(2) = 36` (iii)
After solving Eqs. (i), (ii) and 9iii), we get
`I_(2) = 3` A form b to a.