Ohm’s law in branches — in any series–parallel network, the current through a specific resistor equals the voltage across that resistor divided by its resistance (I = V/R). Does this correctly describe branch current measurement?

Introduction / Context:
Series–parallel circuits can look complicated, but each branch still obeys Ohm’s law locally. Measuring the voltage across one resistor allows you to compute its branch current directly. This question confirms that fundamental relationship for any branch within a linear network.

Given Data / Assumptions:

• Lumped linear elements; constant or slowly varying signals.
• The resistor's two-terminal voltage can be measured.
• Temperature and tolerance effects are negligible for the conceptual calculation.

Concept / Approach:
Ohm’s law states I = V / R for a resistor. Regardless of the complexity of the rest of the network, the local relation between a resistor’s voltage and current is unaffected. Series connections share current; parallel connections share voltage; but within a branch, once V across that element is known, I follows directly via I = V/R.

Step-by-Step Solution:

Verification / Alternative check:
Example: If a 220 Ω branch has 1.70 V across it, I = 1.70 / 220 ≈ 7.73 mA. Measuring with an ammeter in series with that branch should confirm the same value within meter and tolerance limits.

Why Other Options Are Wrong:

• Incorrect: Denies Ohm’s law, which is foundational for linear resistors.
• Only for series-only circuits: Ohm’s law is local and holds in parallel and mixed networks too.
• Only if resistance < 10 kΩ: Magnitude does not invalidate the relation; units and accuracy may change but not the law.

Common Pitfalls:
Confusing the total circuit current with a branch current; mixing up node voltage (common to a parallel group) with element voltage; neglecting meter loading when measuring low-resistance elements.

Final Answer:
Correct — branch current equals the element’s voltage divided by its resistance.