Question

(1) Find the magnitude of current theough 2 Ω.

(2) find the power dissipated in the 6 Ω.

(3) find the voltsge between the junctions

22 3V 3 6V 학 6R 3v

Answer 1

Using Kirchhoff's voltage law in different loop we can find the current in different branches and then using current we can solve other problems.

3 V u B 6V t I, 6 л 3 V I an Let (5, + In) current flowing in the 3r from Gv 3. If I 2 A melt battery La Then it distributed amory two ren's fors and on hamed I & I reap. using Kirchoff's voltagets law Ik loop I (In loop I) 3 (I t I.2) + 2 I 6 - 3 SI, +312 In loop II 3 (I + I) & 6 I 2 3 94 +912 = 3 & then sub with Multiply eq 0 by is I - 3 I using in ew . 2 6-3 40 © 9 - 3 12 IV = 6 Jy = VA Iz=%A a) Hence I current flowing through aut renitor la VA Power dissipated in 6ch rector G2) 6 1x6 366 W to o E-6v voltage b/w the junctions AB: E- ( I, + 12) R 1/2 + %)x3 R=32 AB 6 VAB = uv