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Difficulty: EasyOhm's Law Calculations and Concepts

A technician connects a direct current (DC) indicator lamp across a 10 V10\text{ V} power supply. A digital multimeter connected in series measures a current of 0.5 A0.5\text{ A} flowing through the circuit. What is the electrical resistance of the indicator lamp in ohms (Ω\Omega)?

Answer: 20 ohms

Answer

The resistance of the indicator lamp is 20 Ω20\text{ }\Omega.
According to Ohm's Law, resistance is determined by dividing voltage by current (R=VIR = \frac{V}{I}). Substituting 10 V10\text{ V} and 0.5 A0.5\text{ A} gives R=10 V0.5 A=20 ΩR = \frac{10\text{ V}}{0.5\text{ A}} = 20\text{ }\Omega.

Step-by-Step Solution

1
Identify the given circuit parameters from the problem statement.
Voltage V=10 VV = 10\text{ V}, Current I=0.5 AI = 0.5\text{ A}
Applying Ohm's law requires identifying the known electromotive force and circuit current.
2
Select and arrange the Ohm's law equation to solve for resistance (RR).
R=VIR = \frac{V}{I}
Resistance is equal to the voltage divided by the current flowing through the component.
3
Substitute the measured values into the formula and solve.
R=10 V0.5 A=20 ΩR = \frac{10\text{ V}}{0.5\text{ A}} = 20\text{ }\Omega
Dividing 10 V10\text{ V} by 0.5 A0.5\text{ A} yields the total resistance of 20 Ω20\text{ }\Omega.

Key Concept

Ohm's Law defines the relationship between voltage, current, and resistance in an electrical circuit: V=I×RV = I \times R, which rearranges to R=VIR = \frac{V}{I}.
Estimated Time:45s
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Ohm's Law Calculations and Concepts

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