Question

CAN ANY ONE HELP TO SOLVE THIS PROBLEM SPECIALLY THE LAST PART. FIND i2?

before you come to lab you have to use a similar analysis as the one shown in "Appendix - Sample calculation" to determine a formula for i_2 in circuit 2:

Answer 1

Junction E: $i_2-i_1-i_3 =0$

Loop BCDE: $i_3R_3 - \varepsilon _{2} + i_2R_2=0$

Loop ACDF: $-i_1R_1+i_3R_3 - \varepsilon _{2} +\varepsilon _{1}=0$

Replacing i₁ = i₂ - i₃ in loop ABEF equation

=>$\varepsilon _{1} - (i_2 - i_3)R_1 - i_2R_2 = 0$

=> $\varepsilon _{1} + i_3R_1 - i_2(R_1+R_2) = 0$

Solving this equation with loop BCDE simultaneously

=>$\frac{\varepsilon _{2}-i_2R_2}{R_3} = \frac{i_2(R_1+R_2)-\varepsilon _{1}}{R_1}$

=> $i_{2} = \frac{R_1\varepsilon _{2}+\varepsilon _{1}R_{3}}{R_1R_2+R_1R_3+R_2R_3}$