Power sharing in equal series resistors: A circuit dissipates a total of 12 W across four equal-value resistors in series. How much power does each resistor dissipate?

Introduction / Context:
Knowing how power divides among components helps you select appropriate wattage ratings and avoid overheating. For equal resistors in series, voltage divides equally, and with the same current through each, power splits evenly as well. This question makes you apply those ideas quantitatively to a total power figure.

Given Data / Assumptions:

• Total power P_total = 12 W.
• Four resistors of equal resistance, connected in series.
• Steady DC, ideal components.

Concept / Approach:
In series, the same current flows in all resistors. For equal resistances, each drops the same voltage. Power in each is P_i = I^2 * R (same I and same R), hence equal for all. Therefore, each resistor gets an equal fraction of the total power: P_total / 4.

Step-by-Step Solution:

Verification / Alternative check:
Let each resistor be R and total be 4R. If source voltage is V, then I = V / (4R). Power in one: P1 = I^2 * R = (V^2 / (16R^2)) * R = V^2 / (16R). Total: P_total = 4 * P1 = V^2 / (4R). Solving gives P1 = P_total / 4, matching the simple division used above.

Why Other Options Are Wrong:

• 12 W: That would award the whole power to one resistor.
• 48 W: Exceeds the total power; impossible.
• 8 W: Would imply unequal sharing, which contradicts equal resistances in series.

Common Pitfalls:

• Forgetting that equal R in series split voltage equally, not current (current is already equal).
• Mixing series and parallel intuitions about power distribution.

Final Answer:
3 W