Closing switch adds parallel resistor, reducing terminal voltage and original branch current Steps: - With S open, circuit has battery emf E, internal resistance r, and load R1; total current I = E / (r + R1). - Voltmeter across load reads V = I R1 = E R1 / (r + R1). - Ammeter in R1 branch reads I. - S closed adds R2 in parallel; load R_eq = R1 R2 / (R1 + R2) < R1. - New total current I_total = E / (r + R_eq) > I; new V' = I_total R_eq = E R_eq / (r + R_eq) < V (as x/(r+x) increases with x). - New R1 current I' = V' / R1 < V / R1 = I. Why A is correct: - Parallel path lowers equivalent resistance, increasing total current and internal voltage drop (by Ohm's law, V_terminal = E - I_total r decreases); branch current splits, decreasing ammeter reading. Why the others are wrong: - B: Voltmeter reading decreases, does not increase. - C: Ammeter (branch) reading decreases, does not increase. …