Question

Shown in the figure below is an electrical circuit containing three resistors and two batteries 13 ww Rz G 10 R2 W 12 RA "L Write down the Kirchhoff Junction equation and solve it for Iz in terms of Iz and 13. Write the result here: Write down the Kirchhoff Loop equation for a loop that starts at the lower left corner and follows the perimeter of the circuit diagram clockwise. 0 left Write down the Kirchhoff Loop equation loop that starts at the corner and touches the components 4V, R2, and Rs. 0 = The resistors in the circuit have the following values: • R, -20 • R2 = 40 • Rg=1002 Solve for all the following (some answers may be negative): 1: = Amperes 12 - Amperes 13 - Amperes NOTE: For the equations, put in resistances and currents SYMBOLICALLY using variables like R1,R2,R3 and 11,12,13. Use numerical values of 10 and 4 for the voltages. Submit

Answer 1

13 A B Ww R3 10 R2 С F 4 IZ I. ته E D

a) Kirchoff's junction law states that Total current coming into a juction is equal to total juction coming out of the juction.

Now applying Kirchoff's juction rule at Juction F,

Total inoming current = Total Outgoing current

$I_1+I_3=I_2$

$I_1=I_2-I_3$

b) Applying Kirchoff's voltage rule in EABD,

4+ 13R3 + 10 – 11R1 = 0

c) Applying Kirchoff's voltage rule in EFCD,

$4-I_2R_2-I_1R_1=0$

d) Putting value of $R_1$ ,
RE
, $R_3$ we get the equations as,

$I_1=I_2-I_3$

$4+I_3*10+10-I_1*2=0$

$4-I_2*4-I_1*2=0$

Solving the above three equations we get,

$I_1 = \frac{24}{17}A$

  $I_2 = \frac{5}{17}A$

  $I_3 = -\frac{19}{17}A$