Answer:
[tex]R_{eq} = 100\,\Omega[/tex]
Explanation:
We can solve for the equivalent resistance of the circuit using the formulas for resistors in series and in parallel.
[tex]\displaystyle R_{S} = \sum_i R_i \ \ \ \implies R_{S} = R_1 + R_2 + R_3 + ...[/tex]
[tex]\displaystyle R_P = \dfrac{1}{\displaystyle \sum_i \dfrac{1}{R_i}} \ \ \ \implies R_P = \dfrac{1}{\dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + ...}[/tex]
Plugging the given values into an equation that models this circuit, we get:
[tex]R_{eq} = \dfrac{1}{\dfrac{1}{80\,\Omega + 120\,\Omega} + \dfrac{1}{\dfrac{1}{\dfrac{1}{240\,\Omega} + \dfrac{1}{240\,\Omega}+\dfrac{1}{240\,\Omega}} + 45\,\Omega + 75 \,\Omega}}}[/tex]
[tex]\boxed{R_{eq} = 100\,\Omega}[/tex]
