Question

What is the answer for this

Answer 1

Applying Kirchhoff's second law in closed mesh AFEBA,

9-6 I×1+ I x 1 + 4× = 0 = 0

(current in 42 is given zero)

or

3 = 2I

or

I = 3/2 = 1.5A

now in closed mesh AFEDCBA,

9-3 I(1+1) + IR 6 1.5 × 2 + 1.5R = X

or

6 = 3 + 1.5R

or

R = 3/1.5 = 29

now from point A to point D, (applying Kirchhoff's law for linear part AFED)

Va - I(1 + 1) = VD

or

F VA-VD 1.5 × 2 = 3V