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Zorluk: KolayElectrical Energy and Power

An electric current of 2.0A2.0\,\text{A} flows through a resistor of resistance 5.0Ω5.0\,\Omega for a duration of 10.0seconds10.0\,\text{seconds}. What is the total electrical energy, in Joules, dissipated by the resistor?

Cevap: 200 J

Cevap

The total electrical energy dissipated by the resistor is 200J200\,\text{J}.
According to Joule's law of heating, the electrical energy EE converted into thermal energy when a current II flows through a resistor RR for time tt is given by E=I2RtE = I^2 R t. Substituting I=2.0AI = 2.0\,\text{A}, R=5.0ΩR = 5.0\,\Omega, and t=10.0st = 10.0\,\text{s} gives E=(2.0)2×5.0×10.0=4.0×5.0×10.0=200JE = (2.0)^2 \times 5.0 \times 10.0 = 4.0 \times 5.0 \times 10.0 = 200\,\text{J}.

Adım Adım Çözüm

1
Identify the given physical values from the problem statement.
I=2.0AI = 2.0\,\text{A}, R=5.0ΩR = 5.0\,\Omega, t=10.0st = 10.0\,\text{s}.
These are the necessary parameters to compute electrical energy.
2
State the formula for electrical energy dissipated in a resistor (Joule's law).
E=I2RtE = I^2 R t
Electrical energy is equal to power multiplied by time, where power P=I2RP = I^2 R.
3
Calculate the numerical value of the energy.
E=(2.0)2×5.0×10.0=200JE = (2.0)^2 \times 5.0 \times 10.0 = 200\,\text{J}
Squaring the current gives 4.0A24.0\,\text{A}^2, and multiplying by 5.0Ω5.0\,\Omega and 10.0s10.0\,\text{s} yields 200J200\,\text{J}.

Anahtar Kavram

Joule's Law of Electrical Heating
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