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Question

Difficulty: MediumOhm's Law

A technician is testing a telemetry circuit board and measures a steady direct current of 40 mA40\text{ mA} flowing through a 750 Ω750\ \Omega load resistor. What is the voltage drop across this resistor?

  1. A
    300 V300\text{ V}
  2. B
    18.75 V18.75\text{ V}
  3. 30 V30\text{ V}Answer
  4. D
    3.0 V3.0\text{ V}

Answer

30 V30\text{ V}
According to Ohm's Law (V=I×RV = I \times R), the voltage drop across a component is calculated by multiplying the current flowing through it by its resistance. Converting 40 mA40\text{ mA} to standard amperes gives 0.040 A0.040\text{ A}. Multiplying 0.040 A0.040\text{ A} by 750 Ω750\ \Omega yields a potential difference of 30 V30\text{ V}.

Step-by-Step Solution

1
Convert current from milliamperes (mA) to base amperes (A)
I=40 mA=40×10−3 A=0.040 AI = 40\text{ mA} = 40 \times 10^{-3}\text{ A} = 0.040\text{ A}
Standard electrical calculations using Ohm's Law require base units of amperes, ohms, and volts.
2
Apply Ohm's Law to determine the potential difference (V=I×RV = I \times R)
V=0.040 A×750 Ω=30 VV = 0.040\text{ A} \times 750\ \Omega = 30\text{ V}
Voltage drop across a resistive element is directly proportional to the product of the current and resistance.

Key Concept

Ohm's Law (V=I×RV = I \times R) and metric unit conversion
Estimated Time:45s
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Topics

Ohm's Law

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