Series resistors with a DC source: Three resistors of 1.5 kΩ, 470 Ω, and 3.3 kΩ are connected in series to a 25 V supply. What is the total circuit current (express your choice in common units)?

Introduction:Combining series resistances and applying Ohm's law is a foundational skill for predicting current, voltage drops, and power dissipation. This exercise reinforces unit handling and accurate arithmetic across mixed ohm/kilohm values.

Given Data / Assumptions:

• R1 = 1.5 kΩ, R2 = 470 Ω, R3 = 3.3 kΩ.
• V_source = 25 V DC.
• Ideal wires and negligible extra losses.

Concept / Approach:

For series elements, total resistance is the sum of individual resistances. Then apply I = V / R_total. Converting mixed units (Ω and kΩ) to a single unit avoids arithmetic mistakes and yields a clean milliampere result for readability.

Step-by-Step Solution:

Verification / Alternative check:

Back-calculate expected drops: V1 ≈ 4.742 mA * 1500 Ω ≈ 7.11 V; V2 ≈ 2.24 V; V3 ≈ 15.63 V. Sum ≈ 24.98 V (rounding), matching the 25 V source, which verifies the computation.

Why Other Options Are Wrong:

• 210 mA: Two orders of magnitude too large; would require only ~119 Ω total resistance.
• 5.2 mA: Overestimates due to rounding and ignoring exact totals.
• .007 A (7 mA): Too large; would imply ~3.57 kΩ total, not 5.27 kΩ.
• 2.1 mA: Too small; would imply ~11.9 kΩ total.

Common Pitfalls:

• Mixing kΩ and Ω without conversion.
• Rounding too early; carry sufficient precision until the final step.

Final Answer:

4.7 mA