Question

(a) Determine the currents I1, I2, and I3 in the figure above. Assume R = 12 Ω and the internal resistance of each battery is r = 1.0 Ω.
I1 = ____ A
I2 = ____ A
I3 = ____ A

(b) What is the terminal voltage of the 6.0 V battery?
______ V

Answer 1

for the top loop

12 - 21*i1 + 12 - 11*i2 = 0

>>>>> 21*i1 + 11*i2 = 24

>>>>> i2 = (24/11) - (21/11)*i1    (1)
--------------------------------------------

for the bottom loop
i3=i1-i2

12 - 11*i2 - 6 + 28*i3 = 0

>>>>> 12 - 11*i2 - 6 + 28*(i1-i2) = 0

>>>>> 6 - 39*i2 + 28*i1 = 0     (2)

----------------------------------------------

from 1 and 2 >>>> 6 - 39*((24/11) - (21/11)*i1) +28*i1 = 0

>>>>> i1 = 0.772 A

-------------------------------------------------

from 1 >>>>> i2 = 0.708 A

------------------------------------------------

i3=i1-i2 >>>> i3 = 0.772-0.708

>>>>> i3 = 0.064 A

Answer 2

for the top loop
12+i1*1+i1*8+12-i1*1-i2*10-12i1=0
-4i1-10i2=-12------------(1)

for the bottom loop
i3=i1-i2
12-i2-10i2+12(i1-i2)+(i1-i2)-6+15(i1-i2)=0
12--39i2+28i1=0
28i1-39i2=-12------------(2)

solving 1,2
i1=0.798 A
i2=0.881 A
i3=i1-i2=-0.083 A


terminal voltage of 6V battery=v-ir=6+0.083*1=6.083V