Series circuit with identical resistors: Three 680 Ω resistors are connected in series across a 470 V direct-current source. What is the resulting circuit current (the same through each resistor)?

Introduction / Context:
This problem reinforces application of Ohm's law in a pure series network. When resistors are connected in series, their resistances add directly and exactly the same current flows through each element. By computing the total series resistance and dividing the applied voltage by this total, we obtain the loop current, which is also the current through every resistor in the string.

Given Data / Assumptions:

• Three resistors, each 680 Ω, connected in series.
• Source voltage: 470 V (steady DC).
• Assume ideal components (no internal source resistance, no temperature effects).

Concept / Approach:
For series circuits: R_total = R1 + R2 + R3. The loop current is I = V / R_total. Because series current is identical everywhere, the value computed from total quantities equals the current through each resistor. Unit consistency (volts, ohms, amperes) ensures a correct milliampere result after conversion.

Step-by-Step Solution:

Verification / Alternative check:
Back-calculate voltage using the found current: V ≈ I * R_total ≈ 0.23039 * 2040 ≈ 470 V (within rounding), confirming internal consistency. Individual drops also make sense: V_each ≈ I * 680 ≈ 156.7 V; three drops sum to ≈ 470 V, satisfying KVL.

Why Other Options Are Wrong:

• 69 mA and 23 mA: These imply a much larger equivalent resistance than actually present.
• 690 mA: Would require a far smaller total resistance than 2040 Ω.

Common Pitfalls:

• Forgetting that series resistances add linearly.
• Arithmetic slips when dividing leading to an incorrect order of magnitude.

Final Answer:
230 mA