In a series circuit, what is the effect of an open (failed open) resistor on overall circuit operation? Choose the most accurate statement.

Introduction / Context:
Component failures in series circuits often manifest as opens. Recognizing the circuit-level symptom helps technicians troubleshoot quickly, especially in simple DC networks and many appliance control chains.

Given Data / Assumptions:

• A series path with at least one resistor failed open, resulting in a break in continuity.
• Ideal voltage source with no alternate return paths.
• No parasitic leakage significant enough to sustain current.

Concept / Approach:
A series circuit requires a complete path for current. An open anywhere in that path interrupts charge flow completely. With no current, there is no voltage drop across intact series resistors; instead, nearly the entire source voltage appears across the open because it prevents current and thus acts as a break with effectively infinite resistance.

Step-by-Step Solution:
Model the failed resistor as infinite resistance (R_open → ∞).Compute total resistance: R_total → ∞, so I = V_source / R_total → 0 A.Determine drops: with I ≈ 0, intact resistors have ≈ 0 V drop; the open sees the source potential difference across its terminals.Conclude: no current flows in the entire series loop.

Verification / Alternative check:
Measure with an ammeter in series: reading is 0 A. With a voltmeter across the open component, the reading approaches the source voltage, confirming the diagnosis.

Why Other Options Are Wrong:
Flow around the open: in a series path there is no alternate route.

0 V across the open: incorrect; it generally supports the full source voltage.

Total resistance will decrease: the effective resistance increases toward infinity.

Parallel branches statement: irrelevant to a simple series loop.

Common Pitfalls:
Misinterpreting measurements because of meter loading or stray leakage. Always verify continuity and check for hidden parallel paths before concluding an open.

Final Answer:
No current will flow in the circuit.