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Zorluk: KolayElectrical Power and Watt's Law

A ceramic load resistor in a direct-current (DC) circuit has a resistance of 20 Ω20\ \Omega and carries a steady current of 2 A2\text{ A}. What is the total electrical power dissipated by this resistor?

  1. A
    40 W40\text{ W}
  2. 80 W80\text{ W}Cevap
  3. C
    10 W10\text{ W}
  4. D
    800 W800\text{ W}

Cevap

The total electrical power dissipated by the resistor is 80 W80\text{ W}.
According to Watt's Law, when current (II) and resistance (RR) are known, electrical power is calculated using the formula P=I2RP = I^2 R. Substituting I=2 AI = 2\text{ A} and R=20 ΩR = 20\ \Omega gives P=(2 A)2×20 Ω=4×20=80 WP = (2\text{ A})^2 \times 20\ \Omega = 4 \times 20 = 80\text{ W}.

Adım Adım Çözüm

1
Identify the given electrical values and select the appropriate form of Watt's Law.
Given: current I=2 AI = 2\text{ A}, resistance R=20 ΩR = 20\ \Omega. Formula: P=I2RP = I^2 R.
When current and resistance are known, the power dissipated is calculated using P=I2RP = I^2 R.
2
Square the current value.
I2=(2 A)2=4 A2I^2 = (2\text{ A})^2 = 4\text{ A}^2.
Watt's Law requires current to be squared before multiplying by resistance.
3
Multiply the squared current by the resistance to calculate power in watts.
P=4 A2×20 Ω=80 WP = 4\text{ A}^2 \times 20\ \Omega = 80\text{ W}.
Multiplying amperes squared by ohms yields power in watts.

Anahtar Kavram

Calculating electrical power from current and resistance using Watt's Law (P=I2RP = I^2 R)
Tahmini Süre:45s
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Electrical Power and Watt's Law

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