Use kirchoffs rule only to determine current in each resisto

Use kirchoff\'s rule only to determine current in each resistor determine potential between A & B.

Solution

Let us consider the two loops. Loop-(I) for 12 V battery source.

And loop-(II) for 9 V battery source.

Also let us consider current \'i1\' flows in loop -(I) and current \'i2\' flows in loop-(II).

Now applying kirchoff\'s law in loop-(I) -

3.9*i1 + 1.2*(i1+i2) + 9.8*i1 = 12

=> 14.9*i1 + 1.2*i2 = 12-----------------------------------------------------(a)

For loop-(II), the equation becomes -

6.7*i2 + 1.2*(i2+i1) = 9

=> 1.2*i1 + 7.9*i2 = 9

=> i1 + 6.58*i2 = 7.5

=> i1 = 7.5 - 6.58*i2-----------------------------------------------(b)

Putting the value of i1 from equation (b) to equation (a) -

=> 14.9*(7.5 - 6.58*i2) + 1.2*i2 = 12

=> 96.84*i2 = 99.75

=> i2 = 1.03 A

So, i1 = 7.5 - 6.58*1.03 = 0.72 A

Therefore, current in 3.9 ohm and 9.8 ohm resistor = 0.72 A

Current in 6.7 ohm resistor = 1.03 A

and current in 1.2 ohm resistor = 1.03 A + 0.72 A = 1.75 A

And potential difference between A and B = 9.0 - 6.7*1.03 = 2.1 V