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Difficulty: HardElectrical Power and Watt's Law

An electronic circuit contains a 20 kΩ20\text{ k}\Omega precision resistor that dissipates 180 mW180\text{ mW} of electrical power. What is the DC voltage drop across this resistor?

  1. A
    9 V9\text{ V}
  2. 60 V60\text{ V}Answer
  3. C
    6 V6\text{ V}
  4. D
    3 V3\text{ V}

Answer

The DC voltage drop across the resistor is 60 V60\text{ V}.
According to Watt's Law, electrical power dissipated in a resistive circuit can be expressed as P=V2RP = \frac{V^2}{R}. Solving for voltage yields V=P×RV = \sqrt{P \times R}. Converting the prefix values gives 180 mW=0.18 W180\text{ mW} = 0.18\text{ W} and 20 kΩ=20000 Ω20\text{ k}\Omega = 20{}000\ \Omega. Multiplying these values gives V2=0.18×20000=3600 V2V^2 = 0.18 \times 20{}000 = 3600\text{ V}^2, and taking the square root gives 3600=60 V\sqrt{3600} = 60\text{ V}.

Step-by-Step Solution

1
Convert given values to standard base units (Watts and Ohms)
P=180 mW=180×10−3 W=0.18 WP = 180\text{ mW} = 180 \times 10^{-3}\text{ W} = 0.18\text{ W}; R=20 kΩ=20×103 Ω=20000 ΩR = 20\text{ k}\Omega = 20 \times 10^{3}\ \Omega = 20{}000\ \Omega
Standard electrical formulas require base SI units to prevent magnitude errors.
2
Select and rearrange the appropriate Watt's Law formula relating power, voltage, and resistance
P=V2R  ⟹  V2=P×R  ⟹  V=P×RP = \frac{V^2}{R} \implies V^2 = P \times R \implies V = \sqrt{P \times R}
The known variables are power (PP) and resistance (RR), and the unknown variable is voltage (VV).
3
Calculate the product of power and resistance
V2=(180×10−3 W)×(20×103 Ω)=180×20=3600 V2V^2 = (180 \times 10^{-3}\text{ W}) \times (20 \times 10^{3}\ \Omega) = 180 \times 20 = 3600\text{ V}^2
The metric prefix multipliers 10−310^{-3} and 10310^{3} cancel each other out (10−3×103=110^{-3} \times 10^{3} = 1).
4
Take the square root to determine the voltage
V=3600 V2=60 VV = \sqrt{3600\text{ V}^2} = 60\text{ V}
Taking the square root of V2V^2 yields the electromotive force across the component.

Key Concept

Watt's Law power calculation using P=V2RP = \frac{V^2}{R} and rearranging to solve for voltage with metric prefix conversion.

Alternative Method

First, find the current using P=I2×R  ⟹  I=PR=0.18 W20000 Ω=9×10−6=3×10−3 A=3 mAP = I^2 \times R \implies I = \sqrt{\frac{P}{R}} = \sqrt{\frac{0.18\text{ W}}{20{}000\ \Omega}} = \sqrt{9 \times 10^{-6}} = 3 \times 10^{-3}\text{ A} = 3\text{ mA}. Then apply Ohm's Law to find voltage: V=I×R=(3×10−3 A)×(20×103 Ω)=60 VV = I \times R = (3 \times 10^{-3}\text{ A}) \times (20 \times 10^{3}\ \Omega) = 60\text{ V}.
Estimated Time:1m 30s
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Topics

Electrical Power and Watt's Law

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