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 = 26I1 + 7I2 = 4
Discussion:
9 comments Page 1 of 1.
Sri said:
9 years ago
By Cramers rule.
i2 = |4*4-6*2|/|7*4-6*4|.
= 4/4.
= 1A.
i2 = |4*4-6*2|/|7*4-6*4|.
= 4/4.
= 1A.
(4)
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
then get answer.
6i1+7i2=4
3(4i1+4i2)=3*2
2(6i1+7i2)=4*2 then subtract and finally divided
then get answer.
(3)
Nitin Y. Asangi said:
1 decade 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.
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.
(3)
THIRU said:
1 decade ago
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
6I1+7I2=4----(2)X2
THEN
12I1+12I2=6
12I1+14I2=8
-------------
-2I2=-2
I2=1 AMP
I1=-.5 AMP
(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.
4I1 + 4I2 = 2----(i).
6I1 + 7I2 = 4----(ii).
D=(4*7-6*4)=4.
I2=(4*4-6*2)/D=(16-12)/4=1A.
(2)
Sabari said:
1 decade ago
24i1+24i2=+12
(-) (-) (-)
24i2+28i2=+16
0 -4i2=-4
4i2=4
i2=4/4=1
(-) (-) (-)
24i2+28i2=+16
0 -4i2=-4
4i2=4
i2=4/4=1
(1)
Sanjeev said:
1 decade ago
Simply solve the liner equation.
Akshay said:
1 decade ago
Solve the linear equation for I2 by delta method.
Suji said:
1 decade ago
I think using the equalisation formula, and cancel the unwanted term.
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