Question

The average power dissipated in a pure inductance is

\[\begin{gathered}
  A.{\text{ }}\dfrac{1}{2}L{I^2} \\
  B.{\text{ }}L{I^2} \\
  C.{\text{ }}\dfrac{{L{I^2}}}{4} \\
  D.{\text{ }}zero \\
\end{gathered} \]

Answer 1

Hint:- Pure inductance means the circuit contains only inductance and no resistance and capacitance is present in the circuit. So, we can apply the average power formula that is \[{V_{rms}}{I_{rms}}\cos \phi \].

Complete step-by-step answer:
As we know that Average power formula in a circuit is \[{V_{rms}}{I_{rms}}\cos \phi \] where,

\[{V_{rms}}\] is the root mean square voltage.
\[{I_{rms}}\] is the root mean square current.
And \[\phi \] is the phase difference between the voltage and the current.

And, \[\cos \phi = \dfrac{R}{Z}\] where R is the resistance in the circuit and Z is the net impedance of the circuit.

And we can write root mean square voltage i.e. \[{V_{rms}}\] in terms of root mean square current i.e. \[{I_{rms}}\]as,

\[{V_{rms}} = {I_{rms}}Z\]

So, now putting the value of \[{V_{rms}}\] and \[\cos \phi \] in the formula for average power.

\[{P_{avg}} = {I_{rms}}Z \times {I_{rms}} \times \dfrac{R}{Z} = {I^2}_{rms}R\] (1)

Now as we know that if the circuit is pure inductance than there is no resistance present i.e. R = 0

So, putting the value of r in equation 1.

\[{P_{avg}} = {I^2}_{rms} \times 0 = 0\]

So, the average power dissipated in a pure inductance is zero.

Hence, the correct option will be D.

Note:- Whenever we come up with this type of problem then there is an alternate method to find the inductance in the pure inductance case. As we know that in case of pure inductance phase difference between voltage and current is \[90^\circ \]. So, \[\cos \phi = \cos 90^\circ = 0\]. So, the net average power will be zero.