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Zorluk: OrtaElectrical Power

A direct-current sensor heating element has an internal resistance of 12 Ω12\text{ }\Omega and carries a steady current of 2.5 A2.5\text{ A}. What is the electrical power dissipated by the heating element, in watts?

Cevap: 75 W

Cevap

The electrical power dissipated by the heating element is 75 W75\text{ W}.
Electrical power dissipated by a resistive element is calculated using Watt's Law as P=I2RP = I^2 R. Squaring the current of 2.5 A2.5\text{ A} yields 6.25 A26.25\text{ A}^2, and multiplying by the resistance of 12 Ω12\text{ }\Omega results in 75 W75\text{ W}.

Adım Adım Çözüm

1
Identify the known circuit parameters
I=2.5 AI = 2.5\text{ A} and R=12 ΩR = 12\text{ }\Omega
These are the two direct measurements provided for the resistive load.
2
Select the appropriate power formula
P=I2RP = I^2 R
Watt's Law specifies that power dissipation through a resistance carrying a known current equals the square of the current multiplied by the resistance.
3
Calculate the power dissipation
P=(2.5 A)2×12 Ω=6.25×12=75 WP = (2.5\text{ A})^2 \times 12\text{ }\Omega = 6.25 \times 12 = 75\text{ W}
Squaring 2.52.5 gives 6.256.25, and multiplying 6.256.25 by 1212 produces 7575 watts.

Anahtar Kavram

Calculating electrical power using current and resistance (P=I2RP = I^2 R)
Tahmini Süre:1m 0s
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