Equivalent resistance for resistors in parallel 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:

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

Current:

$\overline{){\mathbf{i}}{\mathbf{=}}\frac{\mathbf{V}}{\mathbf{R}}}$

R_{2} and R_{4} are in series.

R_{24} = 7.5 + 10 = 17.5 Ω

Consider the circuit shown below. Suppose four resistors in this circuit have the values of R_{1} = 13 Ω, R_{2} = 7.5 Ω, R_{3} = 7.7 Ω and R_{4} = 10 Ω, and that the emf of the battery is 18 V.

Find the current through each resistor using the rules for series and parallel resistors

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