# Hi i need help .......any one plz !!! The currents i1 and i2 in the given circuit is .....find?

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kspcr111 | Student

This is simple and is as follows:

see the figure which is in the attachments: according to it

Applying Kirchhoffâ€™s 2nd law for closed loop 1
we get ,

`2-(2*(i1))-(4*((i1)+(i2))) =0` ---------(1)

and now in loop 2

`-1.5 +2*(i2)+(4*((i1)+(i2)))` =0 ----------(2)

Now ,

(1) ==>` 2-(2*(i1))-(4*((i1)+(i2)))` =0

==> `2-(2*(i1))=(4*((i1)+(i2))) ` ------(3)

now applying (3) in (2) we get

=>`-1.5 +2*(i2)+(4*((i1)+(i2)))` =0

=> `-1.5 +2*(i2)+ 2-(2*(i1))` =0

=>` 0.5+ (2*((i2)-(i1)))` =0

=> `(2*((i1)-(i2)))=0.5`

=>`(4*((i1)-(i2))) =1 ` -------------(4)

From (3) and (4) we get on adding

(4*((i1)+(i2)))=2-(2*(i1))

(4*((i1)-(i2))) =1
(+)
-----------------------------
(8*(i1))= 3-(2*(i1))

=> `(10*(i1))= 3`

=>` i1 =0.3`

Now From (4) we get i2

`(4*((i1)-(i2))) =1`

=> `(4*((0.3)-(i2))) =1`

=> `1.2-(4*(i2))=1`

=>` (4*(i2))=0.2`

=> `i2= 0.05`

so, i1 =0.3 & i2= 0.05

:)

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sciencesolve | Student

The given circuit has two loops. Using Kirchoff's current law and Kirchoff's voltage law, yields:

`i1 + i2 = i3` (Kirchoff's current law)

`2 = 2i1 + 4i3` (Kirchoff's voltage law for the left loop)

`1.5 = 2i2 + 4i3 ` (Kirchoff's voltage law for the right loop)

`2 - 1.5 = 2i1 - 2i2` (Kirchoff's voltage law for the entire loop)

Using the equation `i1 + i2 = i3` , yields:

`2i1 + 4(i1+i2) = 2 => 2i1 + 4i1 + 4i2 = 2 => 6i1 + 4i2 = 2 => 3i1 + 2i2 = 1`

`2i2 + 4(i1+i2) = 1.5 = > 4i1 + 6i2 = 1.5`

Multiplying by -3 the equation `3i1 + 2i2 = 1` , yields:

-`9i1 - 6i2 = -3`

Adding the equation `-9i1 - 6i2 = -3` to the equation `4i1 + 6i2 = 1.5` , yields:

`-5i1 = -1.5 => i1 = 1.5/5 => i1 = 0.3` Amps

Substitution of `i1=0.3 ` in equation `3i1 + 2i2 = 1` , gives the following value for i2:

`0.9 + 2i2 = 1 => 2i2 = 0.1 => i2 = 0.1/2 => i2 = 0.05` Amps

Hence, evaluating the currents i1 amd i2, under the given conditions, yields` i1 = 0.3` Amps and `i2 = 0.05` Amps.

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