Kirchhoff's voltage law:

$\overline{){\mathbf{\Sigma}}{\mathbf{V}}{\mathbf{=}}{\mathbf{0}}}$

Voltage:

$\overline{){\mathbf{V}}{\mathbf{=}}{\mathbf{i}}{\mathbf{R}}}$

Let's apply Kirchhoff's loop rule to the top closed loop in a clockwise direction:

$\begin{array}{rcl}{\mathbf{\epsilon}}_{\mathbf{1}}\mathbf{-}{\mathbf{I}}_{\mathbf{2}}{\mathbf{r}}_{\mathbf{1}}\mathbf{-}{\mathbf{I}}_{\mathbf{1}}{\mathbf{R}}_{\mathbf{1}}\mathbf{-}{\mathbf{I}}_{\mathbf{2}}{\mathbf{R}}_{\mathbf{2}}& \mathbf{=}& \mathbf{0}\\ \mathbf{11}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{5}{\mathbf{I}}_{\mathbf{2}}\mathbf{-}\mathbf{6}\mathbf{.}\mathbf{0}{\mathbf{I}}_{\mathbf{1}}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{5}{\mathbf{I}}_{\mathbf{2}}& \mathbf{=}& \mathbf{0}\\ \mathbf{11}\mathbf{-}\mathbf{6}\mathbf{.}\mathbf{0}{\mathbf{I}}_{\mathbf{1}}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{0}{\mathbf{I}}_{\mathbf{2}}& \mathbf{=}& \mathbf{0}\end{array}$

6.0I_{1} + 3.0I_{2} = 11 ...................... (1)

Consider the following circuit of three resistors (R_{1}, R_{2}, and R_{3}), with batteries that have emfs ε_{1} = 11 V and ε_{2} = 44.5 V, and internal resistances r_{1} and r_{2}.

Part A. Find the current I_{1}, in amps.

Part B. Find the current I_{2}, in amps.

Part C. find the current I_{3}, in amps.

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