(a) Determine the currents I 1 {\displaystyle I_{1}} , I 2 {\displaystyle I_{2}} and I 3 {\displaystyle I_{3}} . Assume the internal resistance of each battery is r = 1 Ω {\displaystyle r=1\Omega }
(b) What is the terminal voltage of the 6.0-V battery?
I 1 = I 2 + I 3 {\displaystyle I_{1}=I_{2}+I_{3}}
12.0 V − I 2 ( 12 Ω ) + 12.0 V − I 1 ( 35 Ω ) = 0 {\displaystyle 12.0{\textrm {V}}-I_{2}(12\Omega )+12.0{\textrm {V}}-I_{1}(35\Omega )=0} → 24 = 35 I 1 + 12 I 2 {\displaystyle \rightarrow 24=35I_{1}+12I_{2}}
12.0 V − I 2 ( 12 Ω ) − 6.0 V − I 3 ( 34 Ω ) = 0 {\displaystyle 12.0{\textrm {V}}-I_{2}(12\Omega )-6.0{\textrm {V}}-I_{3}(34\Omega )=0} → 6 = 12 I 2 − 34 I 3 {\displaystyle \rightarrow 6=12I_{2}-34I_{3}}
24 = 35 I 1 + 12 I 2 = 35 ( I 2 + I 3 ) + 12 I 2 = 47 I 2 + 35 I 3 {\displaystyle 24=35I_{1}+12I_{2}=35(I_{2}+I_{3})+12I_{2}=47I_{2}+35I_{3}}
Solve for I 3 {\displaystyle I_{3}} :
6 = 12 I 2 − 34 I 3 → I 2 = 6 + 34 I 3 12 {\displaystyle 6=12I_{2}-34I_{3}\rightarrow I_{2}={\frac {6+34I_{3}}{12}}}
24 = 47 6 + 34 I 3 12 + 35 I 3 {\displaystyle 24=47{\frac {6+34I_{3}}{12}}+35I_{3}}
I 3 = 2.97 mA {\displaystyle I_{3}=2.97{\textrm {mA}}}
then the rest is straightforward
I 2 = 6 + 34 I 3 12 = 0.508 A {\displaystyle I_{2}={\frac {6+34I_{3}}{12}}=0.508{\textrm {A}}}
I 1 = I 2 + I 3 = 0.511 A {\displaystyle I_{1}=I_{2}+I_{3}=0.511{\textrm {A}}}
The terminal voltage at that battery is
6.0 V − I 3 r = 6.0 V − ( 2.97 mA ) ( 1.0 Ω ) = 5.997 V {\displaystyle 6.0{\textrm {V}}-I_{3}r=6.0{\textrm {V}}-(2.97{\textrm {mA}})(1.0\Omega )=5.997{\textrm {V}}}
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and 
. Assume the internal resistance of each battery is 
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