Calculate the total resistance of the circuit.

Okay, let's break down the calculation of the total resistance ((R_o)) step by step. 1. Simplify Series Resistances: * Top Branch: Resistors (R_1) and (R_2) are in series. So, their equivalent resistance (R_{12}) is: $$R_{12} = R_1 + R_2 = 20 \Omega + 40 \Omega = 60 \Omega$$ * Bottom Branch: Resistors (R_3) and (R_4) are in series. So, their equivalent resistance (R_{34}) is: $$R_{34} = R_3 + R_4 = 60 \Omega + 100 \Omega = 160 \Omega$$ * Middle Branch: Resistors (R_6) and (R_7) are in series. So, their equivalent resistance (R_{67}) is: $$R_{67} = R_6 + R_7 = 30 \Omega + 50 \Omega = 80 \Omega$$ * Bottom Branch: Resistors (R_8) and (R_9) are in series. So, their equivalent resistance (R_{89}) is: $$R_{89} = R_8 + R_9 = 60 \Omega + 20 \Omega = 80 \Omega$$ * Bottom Branch: Resistors (R_{10}) and (R_{11}) are in series. So, their equivalent resistance (R_{10,11}) is: $$R_{10,11} = R_{10} + R_{11} = 80 \Omega + 20 \Omega = 100 \Omega$$ 2. Simplify Parallel Resistances (First Set): * Now, (R_{12}) and (R_{34}) are in parallel. The equivalent resistance (R_{a}) is: $$R_{a} = \frac{R_{12} \cdot R_{34}}{R_{12} + R_{34}} = \frac{60 \Omega \cdot 160 \Omega}{60 \Omega + 160 \Omega} = \frac{9600}{220} \approx 43.64 \Omega$$ 3. Simplify Parallel Resistances (Second Set): * Resistors (R_{67}), (R_{89}), (R_{10,11}), (R_{12}) and (R_{13}) are in parallel. To find the equivalent resistance (R_{b}), we use the formula: $$\frac{1}{R_{b}} = \frac{1}{R_{67}} + \frac{1}{R_{89}} + \frac{1}{R_{10,11}} + \frac{1}{R_{12}} + \frac{1}{R_{13}} = \frac{1}{80} + \frac{1}{80} + \frac{1}{100} + \frac{1}{120} + \frac{1}{10}$$ $$\frac{1}{R_{b}} = \frac{3+3}{240} + \frac{1}{100} + \frac{1}{120} + \frac{1}{10}$$ $$\frac{1}{R_{b}} = \frac{6}{240} + \frac{1}{100} + \frac{1}{120} + \frac{1}{10} = \frac{1}{40} + \frac{1}{100} + \frac{1}{120} + \frac{1}{10}$$ $$\frac{1}{R_{b}} = \frac{30+12+10+120}{1200} = \frac{172}{1200} = \frac{43}{300}$$ $$R_{b} = \frac{300}{43} \approx 6.98 \Omega$$ 4. Simplify Series Resistances Again: * Now, we have (R_{a}), (R_5), (R_{b}) and (R_{14}) in series. So, the total equivalent resistance (R_{o}) is: $$R_{o} = R_{a} + R_5 + R_{b} + R_{14} = 43.64 \Omega + 80 \Omega + 6.98 \Omega + 30 \Omega = 160.62 \Omega$$ Therefore, Answer: 160.62 \Omega