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.
   I = I1 +I2I1 = I - I2
By applying voltage law
Point A is at +2V, Point B is junction point. Point C is at -4V and D at -8V.
VAB = ir - E-2-4 = -2I -8I1 6 = 2I + 8I16 = 2I + 8(I-I2) 6= 10I -8I2.....Equation-1-2 +8 = 2I +4I26 =2I +4I2 ...Equation-2Solving 1 and 2 we get I2 = 1.57 A


  • 5
What are you looking for?