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Difficulty: MediumSeries Circuits

An industrial automation control loop is powered by a 48 V48\text{ V} direct current (DC) source connected in series with three resistors: R1=20 ΩR_1 = 20\ \Omega, R2=50 ΩR_2 = 50\ \Omega, and R3=90 ΩR_3 = 90\ \Omega. What is the voltage drop across resistor R2R_2?

  1. A
    48 V48\text{ V}
  2. 15 V15\text{ V}Answer
  3. C
    0.3 V0.3\text{ V}
  4. D
    16 V16\text{ V}

Answer

The voltage drop across the 50 Ω50\ \Omega resistor (R2R_2) is 15 V15\text{ V}.
The correct answer is 15 V15\text{ V}. In a series circuit, the total resistance is the sum of all individual resistors: 20 Ω+50 Ω+90 Ω=160 Ω20\ \Omega + 50\ \Omega + 90\ \Omega = 160\ \Omega. The circuit current is therefore 48 V/160 Ω=0.3 A48\text{ V} / 160\ \Omega = 0.3\text{ A}. Applying Ohm's Law across R2R_2 gives a voltage drop of 0.3 A×50 Ω=15 V0.3\text{ A} \times 50\ \Omega = 15\text{ V}.

Step-by-Step Solution

1
Calculate the total circuit resistance (RTR_T) by summing the three series resistances.
RT=R1+R2+R3=20 Ω+50 Ω+90 Ω=160 ΩR_T = R_1 + R_2 + R_3 = 20\ \Omega + 50\ \Omega + 90\ \Omega = 160\ \Omega
In a series circuit, total resistance is the algebraic sum of all individual load resistances.
2
Determine the total circuit current (ITI_T) using Ohm's Law.
IT=VSRT=48 V160 Ω=0.3 AI_T = \frac{V_S}{R_T} = \frac{48\text{ V}}{160\ \Omega} = 0.3\text{ A}
Current equals total voltage divided by total resistance, and current remains constant through every component in a single series loop.
3
Calculate the specific voltage drop (V2V_2) across resistor R2R_2.
V2=IT×R2=0.3 A×50 Ω=15 VV_2 = I_T \times R_2 = 0.3\text{ A} \times 50\ \Omega = 15\text{ V}
Applying Ohm's Law (V=I×RV = I \times R) to an individual component determines the potential difference dropped across that load.

Key Concept

Series circuit total resistance, current constancy, and individual component voltage drop calculation using Ohm's Law.
Estimated Time:1m 0s
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Topics

Series Circuits

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