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Power dissipated in pure inductance will be:
A. $\dfrac{L{{I}^{2}}}{2}$
B. \[2L{{I}^{2}}\]
C. $\dfrac{L{{I}^{2}}}{4}$
D. zero

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Last updated date: 20th Jun 2024
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Answer
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Hint: Power dissipation is the resultant of the hindrance in the path of the current flow due to the resistance offered in the circuit. The greater the amount of the resistance applied in the circuit higher will be the energy loss. The resistance is due to the resistive property of the circuit.

Complete step-by-step answer:
Three passive components can be included in a circuit. Resistor, inductor, and capacitor. These components are responsible for the hindrance in the flow of the current in the circuit. This hindrance occurs due to the resistance offered in the circuit.
A resistor is a leaner component that follows the rules as per the ohm's law. The inductor and capacitor are the energy-storing devices. When we take an account of a pure inductor circuit.
A circuit of the pure inductor is shown here,
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Power in a circuit is given by,
\[P=IV\cos \theta \]
In the case of an inductor, the voltage leads the current by an angle of $90{}^\circ $
So the power of the circuit is given by,
\[\begin{align}
  & P=IV\cos 90{}^\circ \\
 & \Rightarrow P=0 \\
\end{align}\]
As the value of $cos90{}^\circ$ is zero the energy dissipated in the case of a pure inductor is also zero.

So, the correct answer is “Option D”.

Note: Inductor and capacitor are energy storing devices. In ideal cases, there are no power losses in pure inductor and capacitor circuits. Power loss is due to the resistance offered by the linear passive component (Resistor) or the resistance of the conductor and other components. In the practical world, every electrical component has its internal resistance including the conductor (connecting wires).