Fault/open-branch scenario — in a two-branch parallel network powered by an ideal source, if resistor R2 opens (becomes an open circuit), what happens to the voltage measured across R1?

Introduction / Context:
Troubleshooting parallel networks often involves open-circuit faults in one branch. This question evaluates whether you understand that voltage in a parallel network is common to all branches when driven by an ideal source.

Given Data / Assumptions:

• R1 and R2 are connected in parallel across an ideal voltage source.
• R2 experiences an open-circuit fault (infinite resistance).
• Wiring drops and source internal resistance are negligible.

Concept / Approach:
In parallel, each branch is across the same two nodes. An open in one branch removes current in that branch but does not change the node-to-node voltage set by an ideal source. Therefore, the voltage across R1 remains equal to the source voltage before and after R2 opens.

Step-by-Step Solution:

Verification / Alternative check:
A quick KCL check shows the source now supplies only I_R1. KVL still fixes the node-to-node potential to the source value, leaving V_R1 the same as before.

Why Other Options Are Wrong:

• Increase: No mechanism increases the source-imposed node voltage in an ideal model.
• Decrease: Would require non-ideal source droop or wiring drops not assumed here.
• Equal 0 V: Only true if the source itself is zero or the branch is shorted, neither of which occurs.

Common Pitfalls:
Assuming branch interactions change the common bus voltage; in ideal analysis, opening one branch changes current distribution, not the parallel voltage.

Final Answer:
The voltage across R1 remains unchanged when R2 opens (ideal source).