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Zorluk: OrtaElectrical Power and Watt's Law

A direct-current (DC) avionics heating resistor with a resistance of 200 Ω200\ \Omega dissipates 18 W18\text{ W} of electrical power under steady-state operating conditions. What is the magnitude of the current flowing through the resistor?

  1. A
    90 mA90\text{ mA}
  2. 300 mA300\text{ mA}Cevap
  3. C
    0.30 mA0.30\text{ mA}
  4. D
    60 mA60\text{ mA}

Cevap

300 mA300\text{ mA}
According to Watt's Law, power dissipation in a resistive load is given by P=I2RP = I^2 R. Rearranging to solve for current gives I=PR=18 W200 Ω=0.09 A2=0.3 AI = \sqrt{\frac{P}{R}} = \sqrt{\frac{18\text{ W}}{200\ \Omega}} = \sqrt{0.09\text{ A}^2} = 0.3\text{ A}. Multiplying by 10001{}000 converts the current to 300 mA300\text{ mA}.

Adım Adım Çözüm

1
Select the appropriate Watt's Law formula that relates power (PP), current (II), and resistance (RR).
P=I2RP = I^2 R
The known quantities are electrical power (18 W18\text{ W}) and resistance (200 Ω200\ \Omega), and the target quantity is current (II).
2
Rearrange the formula to solve for the squared current (I2I^2).
I2=PR=18 W200 Ω=0.09 A2I^2 = \frac{P}{R} = \frac{18\text{ W}}{200\ \Omega} = 0.09\text{ A}^2
Isolating the variable term allows direct computation of I2I^2.
3
Take the square root of both sides to calculate the current in amperes (AA).
I=0.09 A2=0.3 AI = \sqrt{0.09\text{ A}^2} = 0.3\text{ A}
Current is the square root of I2I^2.
4
Convert the current from amperes (AA) to milliamperes (mA\text{mA}).
0.3 A×1000 mA/A=300 mA0.3\text{ A} \times 1{}000\text{ mA/A} = 300\text{ mA}
Expressing the value in standard sub-unit milliamperes.

Anahtar Kavram

Watt's Law relates power, current, and resistance through the formula P=I2RP = I^2 R, which can be rearranged as I=PRI = \sqrt{\frac{P}{R}}.
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