Three resistances are connected to form a T-shape as shown in the figure. Then the current in the 4Ω resistor is 

(a) 0.93 A                             (b) 1.42 A                           (c) 2.5 A                    (d) 1.57 A
 

Dear students,
We will solve this using Kirchhoff's first law at junction B.
Let current throug 2ohm is I and 8ohm I₁ and 4ohm I₂.
Then
   $$\begin{gathered} I = I_1 +I_2 \\ {I}_1 = I - I_2 \end{gathered}$$
By applying voltage law
Point A is at +2V, Point B is junction point. Point C is at -4V and D at -8V.
$$\begin{gathered} V_{AB} = \sum ir -\sum E \\ -2-4 = -2I -8I_1 \\ 6 = 2I + 8I_1 \\ 6 = 2I + 8(I-I_2) \\ 6= 10I -8I_2.....Equation-1 \\ -2 +8 = 2I +4I_2 \\ 6 =2I +4I_2 ...Equation-2 \\ Solving 1 and 2 we get I2 = 1.57 A \end{gathered}$$

Regards
Vijaya