Current comparison in parallel: Resistors of 220 Ω, 470 Ω, and 560 Ω are in parallel across the same voltage source. Which resistor carries the least current?

Introduction / Context:
In a parallel network, each branch sees the same voltage. Therefore, the current through each branch depends solely on its resistance via Ohm's law. Recognizing the inverse relationship between current and resistance enables quick qualitative answers without arithmetic.

Given Data / Assumptions:

• Three resistors: 220 Ω, 470 Ω, 560 Ω.
• All are connected in parallel across an identical source voltage.
• Assume ideal conditions, equal applied voltage to each branch.

Concept / Approach:

Ohm's law in each branch: I_branch = V / R_branch. With the same V for all branches, the largest resistance yields the smallest current, and the smallest resistance yields the largest current.

Step-by-Step Solution:

Verification / Alternative check:

Pick any convenient voltage (e.g., 5 V) and compute: I_560 = 5/560 ≈ 8.9 mA; I_470 ≈ 10.6 mA; I_220 ≈ 22.7 mA. The 560 Ω branch is smallest, as expected.

Why Other Options Are Wrong:

220 Ω and 470 Ω have lower resistance, so they must carry more current. 'Impossible to determine' is incorrect because relative currents depend only on R when V is common. 'All equal' contradicts Ohm's law unless resistances are equal.

Common Pitfalls:

Assuming that current splits equally; that only holds for equal resistances.

Final Answer:

560 Ω