We are required to combine parallel and series resistors and determine the current through them.

Equivalent resistance for resistors in parallel:

$\overline{)\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{eq}}}{\mathbf{=}}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{1}}}{\mathbf{+}}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{2}}}{\mathbf{+}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{+}}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{n}}}}$

or for 2 resistors:

$\overline{){{\mathbf{R}}}_{{\mathbf{eq}}}{\mathbf{=}}\frac{{\mathbf{R}}_{\mathbf{1}}{\mathbf{R}}_{\mathbf{2}}}{{\mathbf{R}}_{\mathbf{1}}\mathbf{+}{\mathbf{R}}_{\mathbf{2}}}}$

Equivalent resistance for resistors in series:

$\overline{){{\mathbf{R}}}_{{\mathbf{eq}}}{\mathbf{=}}{{\mathbf{R}}}_{{\mathbf{1}}}{\mathbf{+}}{{\mathbf{R}}}_{{\mathbf{2}}}{\mathbf{+}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{+}}{{\mathbf{R}}}_{{\mathbf{n}}}}$

Consider the circuit shown in the figure below. (Let R = 18.0 Ω).

(a) Find the current in the 18.0-Ω resistor.

(b) Find the potential difference between points a and b.

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