Question

Difficulty: MediumSeries Circuits

An emergency backup lighting circuit is powered by a 48 V48\text{ V} DC supply connected in series with a 30 Ω30\ \Omega current-limiting resistor and two identical LED lamp modules. If a steady current of 0.4 A0.4\text{ A} flows through the circuit during operation, what is the voltage drop across each individual LED lamp module?

  1. A
    12 V12\text{ V}
  2. B
    36 V36\text{ V}
  3. 18 V18\text{ V}Answer
  4. D
    24 V24\text{ V}

Answer

18 V18\text{ V}
In a series circuit, the total voltage of 48 V48\text{ V} is distributed across all components according to Ohm's Law. With 0.4 A0.4\text{ A} flowing through the 30 Ω30\ \Omega resistor, that resistor drops 12 V12\text{ V}. Subtracting 12 V12\text{ V} from the total 48 V48\text{ V} source leaves 36 V36\text{ V} dropped across the two identical LED modules. Because the modules are identical and in series, they share this remaining voltage equally, resulting in 18 V18\text{ V} across each module.

Step-by-Step Solution

1
Calculate the voltage drop across the 30 Ω30\ \Omega series resistor using Ohm's Law (V=I×RV = I \times R).
VR=0.4 A×30 Ω=12 VV_{R} = 0.4\text{ A} \times 30\ \Omega = 12\text{ V}
Current is identical through every component in a series circuit.
2
Determine the remaining voltage available for the two LED modules by applying Kirchhoff's Voltage Law (Vtotal−VRV_{\text{total}} - V_R).
Vremaining=48 V−12 V=36 VV_{\text{remaining}} = 48\text{ V} - 12\text{ V} = 36\text{ V}
The sum of all voltage drops in a series loop must equal the total supply voltage.
3
Divide the remaining voltage equally between the two identical LED modules.
Vmodule=36 V2=18 VV_{\text{module}} = \frac{36\text{ V}}{2} = 18\text{ V}
Identical series components carry the same current and therefore drop equal portions of the remaining voltage.

Key Concept

In a series circuit, current is constant throughout the single loop, and the total supply voltage equals the sum of the individual component voltage drops.
Estimated Time:1m 0s
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