Equivalent resistance for resistors in series:

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

Equivalent resistor for resistors in parallel:

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

Let's label the resistors as follows:

Resistors 1 and 2 are in series.

R_{12} = R_{1} + R_{2} = R + R = 2R

R_{12} and R_{3} are in parallel.

1/R_{123} = 1/2R + 1/R = 3/2R

What is the net resistance of the circuit connected to the battery in (Figure 1)?

Express your answer in terms of *R*.

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