A series–parallel circuit has two resistors R1 and R2 in series forming one branch, and that series branch is connected in parallel with resistor R3 across a DC source. If R1 becomes an open circuit (fails open), what happens to the voltage across R2?

Introduction / Context:
Diagnosing faults in mixed series–parallel networks requires predicting how voltages redistribute when a component fails. An open-circuit failure interrupts current in its path, dramatically altering voltages on that branch.

Given Data / Assumptions:

• R1 and R2 are in series, and this series branch is in parallel with R3.
• R1 fails open (infinite resistance).
• Ideal DC source; no unintended leakage paths.

Concept / Approach:
With R1 open, the R1–R2 branch is broken. No current can flow through R2 because the series path is incomplete. For an ideal resistor, voltage across it is V = I * R. With I = 0 A through R2, its voltage must be 0 V. Therefore, relative to the normal operating condition (nonzero current), the voltage across R2 decreases (it drops to zero).

Step-by-Step Solution:
Model the fault: R1 → ∞ (open circuit).Branch current: I_branch = Vs / (R1 + R2) → 0 A because R1 + R2 → ∞.Voltage across R2: VR2 = I_branch * R2 = 0 A * R2 = 0 V.Conclusion: The voltage across R2 decreases to zero compared to the healthy state.

Verification / Alternative check:
Measure with a voltmeter in practice: after R1 opens, VR2 collapses to 0 V, while the parallel resistor R3 may still carry current and drop nearly the full source voltage across itself.

Why Other Options Are Wrong:
Increase/remain the same: impossible with zero branch current through an ideal resistor.

Cannot tell: the behavior follows directly from V = I * R and an open path (I = 0).

Reverse polarity: no current, hence no defined polarity drop across R2.

Common Pitfalls:
Assuming some voltage must appear across every resistor. In series branches that are opened, ideal resistors see zero current and therefore zero drop.

Final Answer:
decrease