A multimeter is used to take a measurement as shown in the figure above. What will be the measured output?

Question

BlackTom AI · Accepted Answer

The two resistors are placed in series once the meter (in current mode) is connected between the two exposed terminals; the meter in current mode has very low internal resistance and thus completes the circuit. Total series resistance = 150 Ω + 50 Ω = 200 Ω.

Step-by-step calculation:
1. Battery voltage V = 5.00 V.
2. Total resistance Rtot = 150 Ω + 50 Ω = 200 Ω.
3. Circuit current magnitude I = V / Rtot = 5.00 V / 200 Ω = 0.025 A = 25 mA.

Sign determination (why the reading is negative):
- A multimeter displays positive current when current enters its red (mA) input and leaves the black (COM) lead.
- In the shown hookup the current flows from the battery + through the 150 Ω resistor into the meter black lead, through the meter, out the red lead to the top of the 50 Ω resistor and back to the battery −. Thus current actually enters the meter at the black lead and exits at the red lead — opposite the meter’s positive convention. Therefore the meter will show the same magnitude but with a negative sign: −25 mA.

Why other options are incorrect (briefly):
- 33 mA would be 5 V / 150 Ω (ignores the 50 Ω), so it’s wrong because both resistors are in series.
- 100 mA or 25 mA (positive) ignore the sign orientation (25 mA magnitude is correct but sign is negative).
- Voltage answers (±1.25 V, ±3.75 V, ±3.75 V) are irrelevant because the meter is in current mode and measures current, not voltage.
- −100 mA, −33 mA, −3.75 V etc. are mismatched magnitudes/units.

Conclusion: The measured output is −25 mA, which matches the provided correct answer.

A multimeter is used to take a measurement as shown in the figure above. What will be the measured output?Single choice

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