Mixed series string and current through one element: Resistors of 470 Ω, 680 Ω, 1 kΩ, and 1.2 kΩ are in series on a 20 V source. What is the approximate current through the 680 Ω resistor?

Introduction / Context:
In a DC series circuit, the current is identical through every element. Therefore, to find the current through any one resistor, we can compute the total series resistance and apply Ohm's law once for the entire loop. This checks fluency with combining varied resistor values and converting the result to milliamperes.

Given Data / Assumptions:

• Series resistors: 470 Ω, 680 Ω, 1 kΩ (1000 Ω), and 1.2 kΩ (1200 Ω).
• Source voltage: 20 V (DC).
• Ideal conditions (no internal source resistance).

Concept / Approach:
Total series resistance is the arithmetic sum. One loop current flows, so that same current passes through the 680 Ω resistor. Apply I = V / R_total and express the answer in mA for readability.

Step-by-Step Solution:

Verification / Alternative check:
Voltage on the 680 Ω element: V_680 = I * 680 ≈ 0.00597 * 680 ≈ 4.06 V. Summing drops across all four parts returns approximately 20 V, confirming KVL consistency within rounding.

Why Other Options Are Wrong:

• 60 mA and 300 mA: Would require much smaller total resistance than 3.35 kΩ.
• 30 mA: Still too large; would imply R_total ≈ 667 Ω, not the case here.

Common Pitfalls:

• Treating current through one series element as different from the loop current.
• Arithmetic mistakes when summing mixed kΩ and Ω values.

Final Answer:
6 mA