Electric circuits: in a metallic circuit that becomes short-circuited, what happens to the current magnitude assuming an ideal source and negligible resistance in the short?

Introduction / Context:
A short circuit is a low-resistance path that bypasses the intended load. Recognizing its current implications is fundamental to selecting fuses, breakers, and wire gauges for protection.

Given Data / Assumptions:

• Voltage source with internal resistance much smaller than the normal load.
• Short-circuit path of very low resistance.
• Ohm’s law applies: I = V / R_total.

Concept / Approach:
Current magnitude is inversely related to circuit resistance. When the circuit is shorted, the effective resistance plummets toward the source/internal resistance, causing current to surge well beyond the design value and potentially damaging equipment.

Step-by-Step Solution:
Nominal load: I_load = V / (R_source + R_load).Short-circuit: I_short = V / (R_source + R_short) with R_short ≈ 0.Since R_source ≪ R_load, I_short ≫ I_load, leading to a large current flow.

Verification / Alternative check:
Protective devices are rated by prospective short-circuit current (fault current), confirming that short circuits demand interrupting capacities far higher than normal operating currents.

Why Other Options Are Wrong:
Small/Some current: contradicts Ohm’s law under low resistance.
No current: would require an open circuit, not a short.

Common Pitfalls:
Confusing open and short faults; assuming a source can supply infinite current—real sources limit current, but it is still very large relative to normal operation.

Final Answer:
a large current flow