Equivalent resistance for 2 resistors in parallel:

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

Equivalent resistance for resistors in series:

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

**(a)**

R_{1} and R_{2} are in parallel.

R_{12} = (2.00)(5.00)/(2.00+5.00) = 1.43Ω

**(a)** Consider the network of four resistors shown in the diagram, where *R*_{1} = 2.00 Ω, *R*_{2} = 5.00 Ω, *R*_{3} = 1.00 Ω, and *R*_{4} = 7.00 Ω. The resistors are connected to a constant voltage of magnitude *V*. (Figure 1) Find the equivalent resistance *R*_{A} of the resistor network.

**(b)** Two resistors of resistance *R*_{5} = 3.00 Ω and *R*_{6} = 3.00 Ω are added to the network, and an additional resistor of resistance *R*_{7} = 3.00 Ω is connected by a switch, as shown in the diagram..(Figure 2) Find the equivalent resistance *R*_{B} of the new resistor network when the switch is open.

**(c)** Find the equivalent resistance *R*_{C} of the resistor network described in Part B when the switch is closed.

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