1. **Problem Statement:** Calculate the current through the 8-ohm resistor in the given circuit using Thevenin's theorem. Then, find the current if the 6-V battery connections are reversed. 2. **Thevenin's Theorem:** It states that any linear circuit with voltage sources and resistors can be replaced by a single voltage source ($V_{th}$) in series with a single resistor ($R_{th}$) as seen from the terminals of the load resistor (here, the 8-ohm resistor). 3. **Step 1: Remove the 8-ohm resistor** and find the Thevenin equivalent voltage ($V_{th}$) and resistance ($R_{th}$) across points A and B. 4. **Step 2: Find $V_{th}$:** Calculate the open-circuit voltage across A and B. - The circuit has two batteries: 12 V and 6 V. - Using mesh or node analysis, find voltage at A relative to B. 5. **Step 3: Find $R_{th}$:** Turn off all independent voltage sources (replace batteries with short circuits). - Calculate equivalent resistance seen from A and B. 6. **Step 4: Calculate current through 8-ohm resistor:** $$I = \frac{V_{th}}{R_{th} + 8}$$ 7. **Step 5: Reverse 6-V battery connections:** - Repeat steps 2 to 4 with the 6-V battery polarity reversed. --- **Calculations:** - Original circuit: - $R_{th} = 2 + 1 + 6 = 9\ \Omega$ (since 2,1,6 ohm resistors are in series from A to B excluding 8-ohm resistor) - $V_{th} = 12V - 6V = 6V$ (net voltage considering battery polarities) - Current: $$I = \frac{6}{9 + 8} = \frac{6}{17} \approx 0.353\ A$$ - With 6-V battery reversed: - $V_{th} = 12V + 6V = 18V$ - $R_{th} = 9\ \Omega$ (same as before) - Current: $$I = \frac{18}{9 + 8} = \frac{18}{17} \approx 1.059\ A$$ **Final answers:** - Current through 8-ohm resistor originally: $0.353\ A$ - Current with 6-V battery reversed: $1.059\ A$