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Difficulty: HardSeries Circuits

A military vehicle communication module is powered by a 60 V60\text{ V} DC supply connected in series with three resistive components: a fixed input resistor R1=40 ΩR_1 = 40\ \Omega, an unknown matching resistor R2R_2, and a load resistor R3=80 ΩR_3 = 80\ \Omega. If a technician measures a voltage drop of 12 V12\text{ V} across resistor R1R_1, what is the resistance value of R2R_2?

  1. A
    40 Ω40\ \Omega
  2. B
    160 Ω160\ \Omega
  3. 80 Ω80\ \OmegaAnswer
  4. D
    200 Ω200\ \Omega

Answer

80 Ω80\ \Omega
Because current is uniform throughout a single series path, the current through the entire circuit is determined by the known component R1R_1: I=12 V40 Ω=0.3 AI = \frac{12\text{ V}}{40\ \Omega} = 0.3\text{ A}. With a 60 V60\text{ V} supply, the total circuit resistance is RT=60 V0.3 A=200 ΩR_T = \frac{60\text{ V}}{0.3\text{ A}} = 200\ \Omega. Since RT=R1+R2+R3R_T = R_1 + R_2 + R_3, subtracting the known values gives R2=200 Ω−40 Ω−80 Ω=80 ΩR_2 = 200\ \Omega - 40\ \Omega - 80\ \Omega = 80\ \Omega.

Step-by-Step Solution

1
Calculate the common series loop current using Ohm's Law across the first known resistor.
I=V1R1=12 V40 Ω=0.3 AI = \frac{V_1}{R_1} = \frac{12\text{ V}}{40\ \Omega} = 0.3\text{ A}
In a series circuit, the current is identical at all points along the loop.
2
Calculate the total equivalent resistance of the circuit using the total supply voltage and loop current.
RT=VtotalI=60 V0.3 A=200 ΩR_T = \frac{V_{\text{total}}}{I} = \frac{60\text{ V}}{0.3\text{ A}} = 200\ \Omega
Ohm's Law relates the source voltage to the total circuit resistance and total current.
3
Subtract the known series resistances from the total circuit resistance to isolate the unknown resistor.
R2=RT−R1−R3=200 Ω−40 Ω−80 Ω=80 ΩR_2 = R_T - R_1 - R_3 = 200\ \Omega - 40\ \Omega - 80\ \Omega = 80\ \Omega
Total resistance in a series circuit is the algebraic sum of all individual resistances (RT=R1+R2+R3R_T = R_1 + R_2 + R_3).

Key Concept

In a series circuit, current remains constant through every component, and total circuit resistance is the sum of individual resistances (RT=R1+R2+R3R_T = R_1 + R_2 + R_3).

Alternative Method

Find the loop current I=12 V40 Ω=0.3 AI = \frac{12\text{ V}}{40\ \Omega} = 0.3\text{ A}. Then find the voltage drop across R3R_3: V3=0.3 A×80 Ω=24 VV_3 = 0.3\text{ A} \times 80\ \Omega = 24\text{ V}. Using Kirchhoff's Voltage Law, find the voltage drop across R2R_2: V2=60 V−12 V−24 V=24 VV_2 = 60\text{ V} - 12\text{ V} - 24\text{ V} = 24\text{ V}. Finally, calculate R2=V2I=24 V0.3 A=80 ΩR_2 = \frac{V_2}{I} = \frac{24\text{ V}}{0.3\text{ A}} = 80\ \Omega.
Estimated Time:1m 30s
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Series Circuits

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