We need Kirchhoff's junction rule to solve this problem.

Kirchhoff's junction rule states that current **INTO **a junction is always **EQUAL **to current **OUT **of the junction.

That is,

$\overline{){\mathbf{\sum}}{{\mathbf{I}}}_{\mathbf{i}\mathbf{n}}{\mathbf{=}}{\mathbf{\sum}}{{\mathbf{I}}}_{\mathbf{o}\mathbf{u}\mathbf{t}}}$

We'll also use Kirchoff's loop rule, which is written as:

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

We also need to know, from Ohm's law, that:

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

We can get the current through the 5.00-Ω and 8.00-Ω resistors in terms of I_{1}, I_{2}, and I_{3} using the junction rule.

We then apply Kirchhoff's loop rule to the three loops visible in the circuit.

Calculate the three currents I1,I2, and I3 indicated in the circuit diagram shown in the figure (Figure 1) .

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