## Step 1: Find the Equivalent Resistance ### **Analyze the Circuit:** - The 5Ω and 10Ω resistors are in **parallel**. - The result of that combination is inseries** with 8Ω resistor. The result so far is in **parallel** with a 4Ω resistor. - The entire result is in **series** with the top 4Ω resistor. #### **a. Combine 5Ω and 10Ω in parallel:** \[ \frac{1}{R_{A}} = \frac{1}{5} + \frac{1}{10} = \frac{2}{10} + \frac{1}{10} = \frac{3}{10} \] \[ R_{A} = \frac{10}{3} \approx 3.33\, \Omega \] #### **b. Add 8Ω in series:** \[ R_{B} = R_{A} + 8 = 3.33 + 8 = 11.33\, \Omega \] #### **c. Combine with 4Ω in parallel:** \[ \frac{1}{R_{C}} = \frac{1}{11.33} + \frac{1}{4} \] \[ = \frac{1}{11.33} + \frac{1}{4} \approx .0882 + .25 = .3382 \] \[ R_{C} = \frac{1}{.3382} \approx 2.96\, \Omega \] #### **d. Add the top 4Ω resistor in series:** \[ R_{eq} = R_{C} + 4 = 2.96 + 4 = 6.96\, \Omega \] ### **Explanation:** We combined parallel and series resistors step by step to find the total or equivalent resistance. --- ## Step 2: Find Total Current from Battery \[ V = 8\, V,\quad R_{eq} = 6.96\, \Omega \] \[ I_{total} = \frac{V}{R_{eq}} = \frac{8}{6.96} \approx 1.15\, \text{A} \] ### **Explanation:** Using Ohm's Law, the total current flowing from the battery is found by dividing the total voltage by the equivalent resistance. --- ## **Final Answer:** - **Equivalent Resistance:** \( R_{eq} \approx 6.96~\Omega \) - **Total Current from Battery:** \( I_{total} \approx 1.15~\text{A} \)