Faults in series chains — a short circuit across one element in a series network will cause the total circuit current to decrease. Is this claim accurate for an ideal voltage source?

Introduction / Context:
Troubleshooting often hinges on predicting how faults affect current. In a series circuit powered by an ideal voltage source, replacing a finite resistor with a short (near zero ohms) changes the equivalent resistance. This question tests whether you know in which direction the total current moves under such a fault.

Given Data / Assumptions:

• Ideal voltage source of fixed value.
• Series network of resistors.
• Fault: one resistor becomes a short (its resistance → ~0 Ω).

Concept / Approach:
Total series resistance is the sum of individual resistances. If one resistance collapses to nearly zero, the sum decreases. Ohm’s law for the whole circuit is I_total = V_source / R_total. With a smaller denominator (R_total), the current increases, not decreases. Only non-idealities like source current limits, fuses, or saturation would reduce current, but those are separate protective effects, not ideal circuit behavior.

Step-by-Step Solution:

Verification / Alternative check:
Numerical example: V = 12 V, series 3 Ω + 3 Ω. Normal I = 12 / 6 = 2 A. Short one resistor to 0 Ω → R_total = 3 Ω → I = 12 / 3 = 4 A (increased).

Why Other Options Are Wrong:

• Correct: Opposite to Ohm’s law with a fixed source voltage.
• Only correct with current sources: A current source fixes current; voltage then changes across the network, but that is not the scenario here.
• Only correct if other resistors are large: Magnitude changes, not the direction of the change, depend on values.

Common Pitfalls:
Associating the word “short” with “less current.” A short reduces resistance; with a voltage source, current rises and may blow protective devices.

Final Answer:
Incorrect — a short reduces total resistance and increases total current (ideal source).