Step 1: Set up the circuit. The circuit has a 15 volt source in series with a 100 \(\Omega\) resistor (call it \(R_1\)) that leads to a node, call it node A. From node A, a 100 \(\Omega\) resistor (\(R_2\)) runs straight down to the ground rail, and a second 100 \(\Omega\) resistor (\(R_3\)) runs across to node B, where \(R_{Load}\) sits between node B and the ground rail. To find the Thevenin voltage across \(R_{Load}\), take \(R_{Load}\) out of the circuit and find the open circuit voltage that appears across those two terminals.
Step 2: Check which resistors actually carry current. With \(R_{Load}\) removed, node B has only one wire left, the one coming from \(R_3\). Since nothing else connects to node B, no current can flow through \(R_3\). A resistor with no current through it has no voltage drop across it, so the voltage at node B equals the voltage at node A.
Step 3: Find the voltage at node A. The only closed loop left is the 15 volt source, \(R_1\), and \(R_2\), in series. \[ I = \frac{15}{100+100} = 0.075 \text{ A} \] The voltage at node A is the drop across \(R_2\) (measured from node A to ground): \[ V_A = I \times R_2 = 0.075 \times 100 = 7.5 \text{ V} \]
Step 4: Why the other options are wrong. Option (B), 5.0 V, comes from wrongly treating all three 100 \(\Omega\) resistors as one three way divider: \(15 \times \frac{100}{300} = 5\), which ignores that \(R_3\) carries no current. Option (C), 15.0 V, comes from assuming the open circuit means the full source voltage shows up at the load terminals, forgetting that current still flows through \(R_1\) and \(R_2\) and drops some voltage there. Option (D), 10.0 V, comes from wrongly lumping \(R_2\) and \(R_3\) together in series as a single 200 \(\Omega\) path to ground: \(15 \times \frac{200}{300} = 10\), which is not how the circuit is wired.
Final Answer: The Thevenin voltage across \(R_{Load}\) is 7.5 V. \[ \boxed{V_{th} = 7.5 \text{ V}} \]
