Question

Determine the value of the current I1, I2, and I3 following through the circuit in the Figure using Kirchhoff’s law.

Answer 1

Answer:

I1=-0,75A

I2 = I3-I1 = 2.135A-(-0.75A) = 2.885A

I3 = 2.125A

please rate, hope it helps

Explanation:

let's make a loop between 20 and 30ohm

and the second loop in between 20 and 20 ohm

then

for loop 1

V=V

-20V + 20.I3 + 30I1 = 0

20I3 +30I1 = 20

for loop 2

-20V+ 20.I3 -80V + 20.I2 = 0

20I3+20I2 = 100

and

I3=I1+I2

so lets, subtitute

I2= I3-I1 to loop 2 eq

20I3 + 20.(I3 - I1) = 100

40.I3 - 20I1 = 100

let's write

loop 1

20I3 +30I1 = 20

loop 2

40.I3 - 20I1 = 100

using elimination

40I3 + 60I1 = 40

40I3 -20I1 = 100

80I1 = -60

I1 = -3/4 A

to find I3, subtitute it to loop 1 equation

20I3+ 30(-3/4) = 20

20I3 + -90/4 = 20

20I3 = 20 + 90/4

I3 = 1+ (4.5)/4

I3 = 2.125A

I2 = I3-I1 = 2.135A-(-0.75A) = 2.885A