The current magnitude in each of the three branches of the circuit below are 1.25 A, 0.924 A and 0.31 A respectively
According to Ohm's law,
V = I R
V = Voltage
I = Current
R = Resistance
In loop 1,
33 I1 + 76 ( I1 - I2 ) - 48 + 20 ( I1 - I2 ) = 22
129 I1 - 96 I2 = 70 → ( 1 )
In loop 2,
76 ( I1 - I2 ) + 51 I1 + 8.5 + 42 I2 + 20 ( I1 - I2 ) = 0
- 96 I1 + 189 I2 = 56.5
I1 = ( 189 I2 - 56.5 ) / 96 → ( 2 )
Sub ( 2 ) in ( 1 ),
129 ( 189 I2 - 56.5 ) / 96 - 96 I2 = 70
( 24381 I2 - 7288.5 ) / 96 = 96 I2 + 70
24381 I2 - 7288.5 = 9216 I2 + 6720
15165 I2 = 14008.5
I2 = 0.924 A
I1 = ( 189 * 0.924 - 56.5 ) / 96
I1 = 1.25 A
I3 = I1 - I2
I3 = 1.25 - 0.924
I3 = 0.31 A
Therefore, the current magnitude in each of the three branches of the circuit below are:
The given question is incomplete. The complete question is:
Determine the current magnitude and direction of flow in each of the three branches of the circuit below. The circuit is shown in the image attached.
To know more about current magnitude
https://brainly.com/question/8343307
#SPJ4
