Series with a parallel branch — R1 is in series with a branch formed by R2 ∥ R3. If R2 becomes open-circuit (while the source voltage stays the same), what happens to the total current drawn from the source?

Introduction:Series–parallel networks are everywhere in real designs. Predicting how a single element failure (such as a branch opening) affects total current is a vital diagnostic skill for troubleshooting and safety analysis.

Given Data / Assumptions:

• Topology: R1 in series with the parallel combination of R2 and R3.
• Event: R2 opens (infinite resistance), R3 remains unchanged.
• Source voltage is constant, linear resistive elements only.

Concept / Approach:

Equivalent resistance of a parallel pair is R_eq = (R2 * R3) / (R2 + R3). When R2 is normal, R_eq < min(R2, R3). If R2 opens, the branch reduces to only R3, so R_eq_new = R3, which is greater than the previous parallel equivalent. Total resistance R_total = R1 + R_eq therefore increases. With a fixed source voltage, the source current I_total = V_source / R_total must decrease when R_total increases.

Step-by-Step Solution:

Verification / Alternative check:

Example numbers: let R1 = 100 Ω, R2 = 100 Ω, R3 = 100 Ω. Initially R_eq = 50 Ω and R_total = 150 Ω → I = V / 150. After R2 opens, R_eq = 100 Ω and R_total = 200 Ω → current becomes V / 200, evidently smaller.

Why Other Options Are Wrong:

• increases: Would require a decrease in total resistance, which does not occur when a parallel path is removed.
• remains the same: Total resistance changed, so current cannot remain identical for a fixed source voltage.
• cannot tell / oscillates: The DC resistive outcome is deterministic and decreases as shown.

Common Pitfalls:

• Forgetting that removing a parallel branch increases equivalent resistance.
• Confusing series and parallel effects; series adds, parallel reduces resistance.

Final Answer:

decreases