Question

Calculate the value of current $I_1$ in the section of networks shown in figure.

• A. 13 A
• B. 7 A
• C. 10 A
• D. 3 A

Answer 1

The correct option is A 13 A

By kirchhoff's first law (at point A):

$15A=I_2+8A{I}_2=7A$

By kirchhoff's first law (at point B):

$I_2+3A=I_{3}7A+3A=I_3{I}_3=10A$

By kirchhoff's first law (at point D)

$8A=I_4+5A{I}_4=3A$

By kirchhoff's first law (at point C):

$I_3+I_4=I_{1}10A+3A=I_1{I}_1=13A$

Hence option 'A' is correct.