Equivalent resistance for resistors in parallel:

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

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:

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

Power in a circuit:

$\overline{){\mathbf{P}}{\mathbf{=}}{\mathbf{iV}}{\mathbf{=}}{{\mathbf{i}}}^{{\mathbf{2}}}{\mathbf{R}}{\mathbf{=}}\frac{{\mathbf{V}}^{\mathbf{2}}}{\mathbf{R}}}$

Ohm's law:

$\overline{){\mathbf{V}}{\mathbf{=}}{\mathbf{iR}}}$

Label the resistors as below:

R_{4} and R_{5} are in parallel:

R_{45} = (20.0)(20.0)/(20.0 + 20.0) = 10.0Ω

R_{45} and R_{3} are in series:

R_{345} = 10.0 + 10.0 = 20.0Ω

In the circuit shown in the figure, all the resistors are rated at a maximum power of 1.20 W.

What is the maximum emf that the battery can have without burning up any of the resistors?

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Based on our data, we think this problem is relevant for Professor Backus' class at SUNY College of Technology at Delhi.