Parallel resistors with a fixed supply: Two resistors 1 kΩ and 2 kΩ are connected in parallel across a 12 V source. What is the current through the 2 kΩ branch?

Introduction / Context:
Branch currents in parallel circuits are straightforward because each branch experiences the same applied voltage. This is a core concept in Ohm’s law applications and nodal analysis.

Given Data / Assumptions:

• R_branch = 2 kΩ.
• Parallel connection to a 12 V DC source.
• Ideal components, steady state.

Concept / Approach:
For any branch in parallel, I_branch = V / R_branch. No need to compute the total equivalent resistance unless total current is required.

Step-by-Step Solution:
Use Ohm’s law on the 2 kΩ branch: I_2k = 12 V / 2000 Ω = 0.006 A.Convert to milliamps: 0.006 A = 6 mA.

Verification / Alternative check:
Compute the 1 kΩ branch current (12 mA) and note total current would be 18 mA. Power check: P_total = V * I_total = 12 V * 0.018 A = 0.216 W; equals P1 + P2, confirming consistency.

Why Other Options Are Wrong:
4 mA and 8 mA are incorrect divisions of 12 V by 2 kΩ; 12 mA corresponds to the 1 kΩ branch, not the 2 kΩ branch.

Common Pitfalls:
Accidentally using series current rules or confusing branch labels, leading to swapping the 1 kΩ and 2 kΩ results.

Final Answer:
6 mA