In a DC circuit powered by an 18 V battery, the ammeter initially reads 40 mA but later drops to 20 mA. Assuming an ideal constant-voltage source, by how many volts has the supply voltage changed, if at all?

Introduction / Context:
In basic circuit analysis using Ohm's law, it is critical to distinguish between constant-voltage sources and changes in circuit current caused by resistance variations. Here, the source is specified as an 18 V battery, and the question probes whether a change in current necessarily implies a change in supply voltage.

Given Data / Assumptions:

• Supply: ideal 18 V battery (constant-voltage assumption).
• Initial current: 40 mA.
• Later current: 20 mA.
• No explicit change to the source is mentioned; measurement is of circuit current only.

Concept / Approach:

Ohm's law: V = I * R. With an ideal constant-voltage source, changes in current reflect changes in total circuit resistance or load conditions, not a change in source voltage. Unless the problem states that the battery voltage has drooped or been adjusted, we treat the supply voltage as unchanged.

Step-by-Step Solution:

Verification / Alternative check:

If the source had changed, the problem would explicitly indicate a new supply value. Since it does not, the standard assumption in DC theory exercises is a fixed source. Current variation alone does not imply voltage variation.

Why Other Options Are Wrong:

9 V or 900 mV suggest partial voltage sag without evidence. 18 V implies the supply went to zero, which is not stated.

Common Pitfalls:

Assuming any current change must come from a changing source; forgetting that resistance variations (or load changes) alter current for a fixed voltage.

Final Answer:

0 V