Four 8 Ω speakers in parallel on a 12 V audio amplifier: What total power must the amplifier be able to deliver at the speakers' terminals (maximum)?

Introduction / Context:
Loudspeaker loads are often paralleled to share power. The amplifier must supply both the required voltage and the current set by the total load. This question tests combining parallel resistances and using the power relation P = V^2 / R.

Given Data / Assumptions:

• Number of speakers: 4 (identical, 8 Ω each).
• Connection: parallel across the same 12 V maximum output.
• Assume speakers behave as resistive loads at the test frequency.

Concept / Approach:
For N equal resistors R in parallel, R_total = R / N. With total resistance known, input power at the load is P_total = V^2 / R_total.

Step-by-Step Solution:

Verification / Alternative check:
Per-speaker power at 12 V: P_each = V^2 / 8 = 144 / 8 = 18 W; times 4 speakers = 72 W. Matches the total calculation.

Why Other Options Are Wrong:

• 48 W or 18 W: Represent partial or per-branch power, not the total parallel load.
• 1.5 W: Orders of magnitude too low for 12 V across low-ohmic loads.

Common Pitfalls:

• Confusing series vs. parallel combination when counting speakers.
• Forgetting that total power equals the sum of branch powers in parallel.

Final Answer:
72 W