Four parallel resistors with values R, 2R, 3R, and 4R: If 15 mA flows through the branch R, what is the total current drawn by the network?

Introduction / Context:
In parallel circuits, all branches share the same voltage. Therefore, currents divide inversely with resistance. This problem assesses current division for branches with simple multiples of a base resistance R.

Given Data / Assumptions:

• Four branches with resistances R, 2R, 3R, and 4R.
• Branch current through R is I_R = 15 mA.
• All branches share a common voltage V.

Concept / Approach:
Let V be the common branch voltage. Then I_branch = V / R_branch. Knowing I_R = V / R, we can express other currents as fractions of I_R: I_2R = I_R / 2, I_3R = I_R / 3, I_4R = I_R / 4. Total current is the sum of individual branch currents.

Step-by-Step Solution:

Verification / Alternative check:
Equivalent conductance approach: G_total = (1/R)(1 + 1/2 + 1/3 + 1/4) multiplied by V gives identical total current when V = I_R * R. Summation 1 + 0.5 + 0.333... + 0.25 = 2.0833..., and 15 mA times 2.0833... ≈ 31.25 mA.

Why Other Options Are Wrong:

• 60 mA and 135 mA: Overestimates; would require additional branches or higher voltage.
• 15 mA: Only one branch current, not the total.

Common Pitfalls:

• Treating currents as proportional to resistance instead of the inverse.
• Forgetting to convert fractional results precisely (eighths and thirds).

Final Answer:
31.25 mA