Power dissipation is maximum in
A. Pure capacitive circuit
B. Pure inductive circuit
C. Pure resistive circuit
D. None of these

Hint:We are going to calculate the average power dissipated in each of the circuits(Pure inductive, pure capacitive, pure resistive). To calculate that, let’s understand the formula first.

Formula used:
Average Power dissipated over one complete cycle in an A.C circuit is given by,
\[{{P}_{avg}}={{V}_{r.m.s}}{{I}_{r.m.s}}\cos \phi \]
Where \[\phi \] is the phase difference between the current and the voltage, \[{{V}_{r.m.s}}\] is the root mean square value of the voltage and \[{{I}_{r.m.s}}\] is the root mean square value of the current.

Complete step by step answer:
In the pure capacitive circuit, the current flowing through the capacitor leads the voltage by an angle of \[{{90}^{\circ }}\], which means,
\[\phi ={{90}^{\circ }} \\
\Rightarrow \cos \phi =\cos {{90}^{\circ }}=0 \]
Therefore, power dissipated in pure capacitive circuit is \[{{V}_{r.m.s}}\times {{I}_{r.m.s}}\times 0=0\]

In a pure inductive circuit, the current in the pure inductive A.C circuit lags the voltage by an angle of \[{{90}^{\circ }}\], which means,
\[\phi ={{90}^{\circ }} \\
\Rightarrow \cos \phi =\cos {{90}^{\circ }}=0 \]
Therefore, power dissipated in pure inductive circuit is also \[{{V}_{r.m.s}}\times {{I}_{r.m.s}}\times 0=0\]

In a pure resistive circuit, the phase angle between current and voltage is\[{{0}^{\circ }}\], which means, they are in phase with each other (\[\phi ={{0}^{\circ }}\]). And maximum value of \[\cos \phi =1\] which further suggests that maximum power dissipation occurs in a pure resistive circuit.
\[\therefore \cos \phi =\cos {{0}^{\circ }}=1\]
Therefore, power dissipated in pure resistive circuit is also \[{{V}_{r.m.s}}\times {{I}_{r.m.s}}\times 1={{V}_{r.m.s}}\times {{I}_{r.m.s}}\].

Hence, the correct answer is option C.

Note: In A.C, we define, \[{{V}_{r.m.s}}\] and \[{{I}_{r.m.s}}\] for convenience to calculate power and they both are constant all the time. In A.C circuits, voltage and current values oscillates which makes the instantaneous power oscillate as well and to give power a one value, we use average value of power and that’s what we used in the question. Same goes the explanation for the usage of \[{{V}_{r.m.s}}\] and \[{{I}_{r.m.s}}\].