(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
for the top loop
12 - 21*i1 + 12 - 11*i2 = 0
>>>>> 21*i1 + 11*i2 = 24
>>>>> i2
= (24/11) - (21/11)*i1 (1)
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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)
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from 1 and 2 >>>> 6 - 39*((24/11) - (21/11)*i1) +28*i1 = 0
>>>>> i1 = 0.772 A
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from 1 >>>>> i2 = 0.708 A
------------------------------------------------
i3=i1-i2 >>>> i3 = 0.772-0.708
>>>>> i3 = 0.064 A
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