Question

please answer this question ...​

Answer 1

Answer:

Applying Kirchhoff's 2nd law for closed circuit AFEBA

24 + 2I(1) + 6I(2) = 27

2I(1) + 6I(2) = 3. -----[1]

Applying Kirchhoff's 2nd law for closed circuit BCDEB

4I(3) + 6I(2) = 23. -----[2]

According to Kirchhoff's 1st law

I(1) + I(3) = I(2)

Now from equation

-----[1]. 2I(1) + 6I(1) + 6I(3) = 3

8I(1) + 6I(3) = 3 ------------[3]

-----[2]. 4I(1) + 6I(1) + 6I(3) = 27

6I(1) + 10I(3) = 27 --------[4]

Multiplying equation [3] by 3

It will be.

24I(1) + 18I(3) = 9 --------[5]

Multiplying equation [4] by 4

It will be.

24I(1) + 18I(3) = 108 --------[6]

Now from equations [5] & [6]

( Substracting equa. [5] from equa.[6] ).

I(3) = 9/2

Now putting the value of I(3) in equa. [3]

I(1) = -3

Now putting the value of I(1) in Kirchhoff's first law

i.e. I(1) + I(3) = I(2)

I(2) = 3/2

Hence , the value of I(1) = -3 , I(2) = 3/2 and I(3) = 9/2.

If it helps u than plz mark it brainlist.