Maximum total current from power ratings: a series circuit has three resistors rated 1/8 W, 1/4 W, and 1/2 W with a total resistance of 1200 Ω. If each operates at its own maximum power simultaneously, what is the total circuit current?

Introduction / Context:
In a series circuit, the current is the same through every component. If each resistor is just at its own maximum power rating, the common current must be such that the sum of individual resistor power ratings equals the total power dissipated by the series string. This requires combining power ratings and total resistance to solve for the allowable current.

Given Data / Assumptions:

• Resistor power ratings: 1/8 W (0.125 W), 1/4 W (0.25 W), 1/2 W (0.5 W).
• Total resistance R_total = 1200 Ω (sum of the three resistors).
• Series connection: same current I flows through all.

Concept / Approach:

Total power in series with current I is P_total = I^2 * R_total. If each resistor is at its own maximum power at the same current, the total allowable power equals the sum of the individual power ratings. Therefore, set I^2 * R_total = P1 + P2 + P3 and solve for I.

Step-by-Step Solution:

Verification / Alternative check:

Power consistency: I^2 * R_total = 0.027^2 * 1200 ≈ 0.875 W, matching the sum of ratings, so each resistor is at its limit simultaneously only if their resistances are proportioned accordingly; the problem frames it as a maximum-total-current scenario using the sum of ratings and total R.

Why Other Options Are Wrong:

2.7 mA is too small by 10×; 19 mA is insufficient to reach the combined rating; 190 mA would greatly exceed power ratings and overheat components.

Common Pitfalls:

Trying to use individual resistor limits without total R; forgetting that P_total in a series path depends on I^2 times the sum resistance.

Final Answer:

27 mA