1. **Problem Statement:** Use Thevenin's theorem to calculate the potential difference (p.d.) across terminals A and B in the given circuit. 2. **Thevenin's Theorem Overview:** Thevenin's theorem states that any linear electrical network with voltage sources and resistors can be replaced at terminals A-B by an equivalent voltage source $V_{th}$ in series with a resistance $R_{th}$. 3. **Step 1: Identify the circuit elements:** - Branch 1: 6 \(\Omega\) resistor and 6 V battery - Branch 2: 6 \(\Omega\) resistor and 6 V battery - Branch 3: 3 \(\Omega\) resistor and 4.5 V battery - Branch 4: 3 \(\Omega\) resistor only 4. **Step 2: Find Thevenin voltage $V_{th}$:** Calculate the open-circuit voltage across terminals A and B. Since the branches are connected in parallel between A and B, the voltage across A-B is the voltage of the branch with the highest voltage considering polarity and resistor drops. Calculate voltage drops and battery voltages for each branch: - Branch 1 voltage: 6 V (battery) - voltage drop across 6 \(\Omega\) resistor (no current since open circuit) = 6 V - Branch 2 voltage: similarly 6 V - Branch 3 voltage: 4.5 V - Branch 4 voltage: 0 V (no battery) Since no current flows in open circuit, voltage across A-B is the voltage of the batteries in parallel, which is 6 V. Thus, $$V_{th} = 6\text{ V}$$ 5. **Step 3: Find Thevenin resistance $R_{th}$:** Deactivate all independent voltage sources (replace batteries with short circuits). The resistors are now: - Branch 1: 6 \(\Omega\) - Branch 2: 6 \(\Omega\) - Branch 3: 3 \(\Omega\) - Branch 4: 3 \(\Omega\) All four resistors are in parallel between terminals A and B. Calculate equivalent resistance: $$\frac{1}{R_{th}} = \frac{1}{6} + \frac{1}{6} + \frac{1}{3} + \frac{1}{3} = \frac{1}{6} + \frac{1}{6} + \frac{2}{6} + \frac{2}{6} = \frac{6}{6} = 1$$ So, $$R_{th} = 1\ \Omega$$ 6. **Step 4: Calculate p.d. across terminals A and B:** Since the circuit is open at terminals A and B, the p.d. is simply the Thevenin voltage: $$V_{AB} = V_{th} = 6\text{ V}$$ **Final answer:** The potential difference across terminals A and B is **6 V**.