Equivalent resistance for resistors in parallel:

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

Equivalent resistance for resistors in series:

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

**1)**

R_{2}, R_{3} and R_{4} are in parallel:

R_{234} = (1/R_{2} + 1/R_{3} + 1/R_{4})^{-1} = (1/2 + 1/2 + 1/4) = 0.8Ω

R_{1} and R_{234} are in series:

The battery in the figure below has negligible internal resistance.

Find the current in each resistor.

1)A (3 Ω resistor)

2)A (4 Ω resistor)

3)A (vertical 2 Ω resistor)

4) A (diagonal 2 Ω resistor)

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