Kirchhoff's loop rule:

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

Kirchhoff's junction rule:

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

We'll assign current directions to the diagram. The junctions are already labeled.

**(a)**

At junction c:

i_{1} = i_{2} + i_{3}

In the circuit, we have 3 unknowns (the currents only). We need an expression to substitute for one of the currents so that we remain with only 2 unknowns when working out Kirchhoff's loop rule.

Solving for i_{2}:

i_{2} = i_{1} - i_{3}

We'll proceed to Kirchoff's loop rule. With only two unknowns, we'll consider any two loops. We have three loops: Left, right, and outer loop.

Using Kirchhoff’s rules (and given ε_{1} = 70.7 V, ε_{2} = 61.4 V, ε_{3} = 79.9 V) calculate the following:

(a) Find the current in each resistor shown in the figure above.

(b) Find the potential difference between points c and f.

(c) Which point is at higher potential?

a. c

b. f

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