# Electrical Engineering - Branch, Loop and Node Analyses - Discussion

Discussion Forum : Branch, Loop and Node Analyses - General Questions (Q.No. 5)
5.

Find I2.

4I1 + 4I2 = 2
6I1 + 7I2 = 4
1 A
–1 A
100 mA
–100 mA
Explanation:
No answer description is available. Let's discuss.
Discussion:
9 comments Page 1 of 1.

Simply solve the liner equation.

Solve the linear equation for I2 by delta method.

4I1+4I2=2----(1)X3
6I1+7I2=4----(2)X2
THEN

12I1+12I2=6
12I1+14I2=8
-------------
-2I2=-2
I2=1 AMP
I1=-.5 AMP
(1)

24i1+24i2=+12
(-) (-) (-)
24i2+28i2=+16

0 -4i2=-4

4i2=4

i2=4/4=1

Gohar rahman said:   1 decade ago
4i1+4i2=2
6i1+7i2=4
3(4i1+4i2)=3*2
2(6i1+7i2)=4*2 then subtract and finally divided
(2)

I think using the equalisation formula, and cancel the unwanted term.

Nitin Y. Asangi said:   10 years ago
4I1 + 4I2 = 2----(i).
6I1 + 7I2 = 4----(ii).

Multiply equation (i) by 6 & equation (ii) by 4,

24I1 + 24I2 = 12----(iii).
24I1 + 28I2 = 16----(iv).

Subtracting equation (iii) & (iv),

24I1 + 24I2 = 12.
-24I1 - 28I2 = -16.

-4I2 = -4.
I2 = 1.
(1)

Sri said:   8 years ago
By Cramers rule.

i2 = |4*4-6*2|/|7*4-6*4|.
= 4/4.
= 1A.
(2)

Mahesha B said:   7 years ago
Matirix mathod.

4I1 + 4I2 = 2----(i).
6I1 + 7I2 = 4----(ii).

D=(4*7-6*4)=4.
I2=(4*4-6*2)/D=(16-12)/4=1A.
(1)