Parallel circuits and current sharing: assess the claim that the branch with the largest resistance in a parallel network carries the largest current when driven by the same voltage.

Introduction / Context:
Current division in parallel networks determines how currents split among branches. A common misconception is that a larger resistance somehow draws more current. This question challenges that misconception and reinforces the proportional relationships that govern current sharing.

Given Data / Assumptions:

• Ideal constant-voltage source supplying a set of resistors in parallel.
• Ohmic resistors with fixed values.
• Temperature and nonlinearity effects ignored.

Concept / Approach:
By Ohm law, I = V / R for each branch at the same voltage V. Current is inversely proportional to resistance. Therefore the branch with the smallest resistance carries the largest current, and the branch with the largest resistance carries the smallest current. The current-divider relationship formalizes this inverse proportionality.

Step-by-Step Solution:

Verification / Alternative check:
Use the current-divider form for two branches: I_through_R1 = I_total * (R2 / (R1 + R2)). The numerator contains the other branch resistance, again showing inverse proportionality.

Why Other Options Are Wrong:
Correct: conflicts with inverse relation.
Holds only for current sources: even with an ideal current source, branch currents still split inversely to resistance when branches are in parallel across the same voltage.

Common Pitfalls:
Assuming current spreads equally; mixing up series and parallel intuition; overlooking that voltage is common across branches in parallel.

Final Answer:
Incorrect