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Difficulty: MediumOhm's Law Calculations and Concepts

A direct current (DC) emergency lighting circuit operates from a 28 V28\text{ V} power supply. An in-line ammeter measures a steady current flow of 350 mA350\text{ mA} through the circuit. What is the total electrical resistance of the lighting circuit?

  1. A
    0.08 Ω0.08\text{ }\Omega
  2. B
    9.8 Ω9.8\text{ }\Omega
  3. C
    12.5 Ω12.5\text{ }\Omega
  4. 80 Ω80\text{ }\OmegaAnswer

Answer

The total electrical resistance of the lighting circuit is 80 Ω80\text{ }\Omega.
According to Ohm's law, resistance is found by dividing voltage by current (R=V/IR = V / I). Converting 350 mA350\text{ mA} into amperes yields 0.35 A0.35\text{ A}. Dividing 28 V28\text{ V} by 0.35 A0.35\text{ A} gives the correct value of 80 Ω80\text{ }\Omega.

Step-by-Step Solution

1
Convert the current measurement from milliamperes to amperes.
I=350 mA=3501000 A=0.35 AI = 350\text{ mA} = \frac{350}{1000}\text{ A} = 0.35\text{ A}
Standard Ohm's law calculations require current to be in base SI units (amperes) to produce resistance in ohms.
2
Rearrange Ohm's law (V=I×RV = I \times R) to solve for resistance (RR).
R=VIR = \frac{V}{I}
Resistance equals the electromotive force (voltage) across the circuit divided by the resulting current flow.
3
Substitute the known voltage and converted current values into the rearranged formula.
R=28 V0.35 A=280035 Ω=80 ΩR = \frac{28\text{ V}}{0.35\text{ A}} = \frac{2800}{35}\text{ }\Omega = 80\text{ }\Omega
Dividing 28 V28\text{ V} by 0.35 A0.35\text{ A} determines the load resistance in ohms.

Key Concept

Ohm's Law (V=I×RV = I \times R) defines the mathematical relationship between voltage, current, and resistance in DC circuits, requiring proper metric unit conversion before calculation.
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Ohm's Law Calculations and Concepts

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