(a) Let I 1 be the branch current though R 1 , I 2 be the branch current through

Question

(a) Let I1 be the branch current though R1, I2 be the branch current through R2, and I3 be the branch current through R3. Write Kirchhoff's loop rule relation for a loop that travels through battery 1, resistor 1, and resistor 3. (Use a clockwise current loop when entering your answer. Use any variable or symbol stated above as necessary.)
= 0

(b) Write Kirchhoff's loop rule relation for a loop that travels through battery 2, resistor 2, and resistor 3. (Use a clockwise current loop when entering your answer. Use any variable or symbol stated above as necessary.)
= 0

(c) Apply Kirchhoff's junction rule to the junction at A to get a relation between the three branch currents. (Use any variable or symbol stated above as necessary.)
I1 =  

(d) You should now have three equations and three unknowns (I1, I2, and I3). Solve for the three branch currents.

Explanation / Answer

Given

R1 = 2,600 , R2 = 1,600 , R3 = 4,700 , and R4 = 5,900 and

e1 = 1.5 V and e2 = 3.0 V

here we have two loops Ll ,Lr

applying the kirchhoff ruel for the loop Ll

   e1 -i1R1-i3R3 = 0 -----------(1)

and for the loop Lr

   -e2 + i3R3 -i2R2 = 0 -------------(2)

   -3+ 4700 i3 - 1600 i2 = 0
  
   4700 i3 - 1600 i2 = 3

   1600 i2 - 4700i3 = -3

kirchhoff current rule for the juction A

   i1 = i2+i3 --------------(3)

from (1)and (2)

adding these

   e1-e2- i1R1-i2R2 = 0

   i1R1+i2R2 = e1-e2

   i1R1+i2R2 = 1.5-3 = -1.5

   2600 i1 +1600i2 = -1.5

i1 = i2+i3

   2600(i2+i3) +1600 i2 = -1.5
  
   4200 i2 + 2600 i3 = -1.5

solving for i2,i3 , i2 = -0.62133 mA , i3 = -0.4267 mA

and i1 = -1.04803 mA

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