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Zorluk: OrtaElectrical Power

An electronics technician is evaluating four separate direct-current (DC) loads installed in an automated weather observatory. Match each DC load description with its calculated electrical power dissipation.

  • Telemetry data logger operating at 24 V24\text{ V} and drawing a steady current of 0.25 A0.25\text{ A}6 W6\text{ W}
  • Antenna de-icing heating element with a resistance of 5 Ω5\text{ }\Omega carrying a current of 4 A4\text{ A}80 W80\text{ W}
  • Warning beacon connected across a 36 V36\text{ V} source with an internal resistance of 18 Ω18\text{ }\Omega72 W72\text{ W}
  • Ventilation blower motor operating at 48 V48\text{ V} and drawing a current of 2.5 A2.5\text{ A}120 W120\text{ W}

Cevap

Telemetry data logger (24 V24\text{ V}, 0.25 A0.25\text{ A}) matches 6 W6\text{ W}; Antenna de-icing heating element (4 A4\text{ A}, 5 Ω5\text{ }\Omega) matches 80 W80\text{ W}; Warning beacon (36 V36\text{ V}, 18 Ω18\text{ }\Omega) matches 72 W72\text{ W}; Ventilation blower motor (48 V48\text{ V}, 2.5 A2.5\text{ A}) matches 120 W120\text{ W}.
Each electrical load is correctly paired by selecting and computing the appropriate power equation based on the given electrical parameters: P=V×IP = V \times I for voltage and current, P=I2×RP = I^2 \times R for current and resistance, and P=V2RP = \frac{V^2}{R} for voltage and resistance.

Adım Adım Çözüm

1
Calculate the power dissipation for the telemetry data logger using given voltage and current.
P=V×I=24 V×0.25 A=6 WP = V \times I = 24\text{ V} \times 0.25\text{ A} = 6\text{ W}
When voltage (VV) and current (II) are given, power is the product of voltage and current.
2
Calculate the power dissipation for the antenna de-icing heating element using current and resistance.
P=I2×R=(4 A)2×5 Ω=16×5=80 WP = I^2 \times R = (4\text{ A})^2 \times 5\text{ }\Omega = 16 \times 5 = 80\text{ W}
When current (II) and resistance (RR) are given, power equals the square of the current multiplied by resistance.
3
Calculate the power dissipation for the warning beacon using voltage and resistance.
P=V2R=(36 V)218 Ω=129618=72 WP = \frac{V^2}{R} = \frac{(36\text{ V})^2}{18\text{ }\Omega} = \frac{1296}{18} = 72\text{ W}
When voltage (VV) and resistance (RR) are given, power equals the square of the voltage divided by resistance.
4
Calculate the power dissipation for the ventilation blower motor using voltage and current.
P=V×I=48 V×2.5 A=120 WP = V \times I = 48\text{ V} \times 2.5\text{ A} = 120\text{ W}
Multiplying the operating voltage by the operating current yields the total power consumption in watts.

Anahtar Kavram

Electrical power in direct-current (DC) circuits represents the rate at which electrical energy is converted into other energy forms (such as heat, light, or mechanical motion). The three standard formulas derived from Ohm's Law and Watt's Law are P=V×IP = V \times I, P=I2×RP = I^2 \times R, and P=V2RP = \frac{V^2}{R}.
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