In a AC circuit the voltage and current are described by \[V = 200\sin \left( {319t - \dfrac{\pi }{6}} \right){\text{ volts}}\] and $I = 50\sin \left( {314t + \dfrac{\pi }{6}} \right){\text{ mA}}$ respectively. The average power dissipated in the circuit is
A. 2.5 watts
B. 5.0 watts
C. 10.0 watts
D. 50.0 watts

Hint: The average power dissipated in an electrical circuit is given as half of the product of peak voltage and peak current with the cosine of the phase difference between current and voltage. From the given expressions for current and voltage, we can get their peak values and phase difference, using them we can easily calculate the average power dissipated in the given circuit.

Formula used:
In an AC circuit, the current can be defined as
$I = {I_0}\sin \left( {\omega t + {\theta _1}} \right)$.............…(i)
Similarly, the voltage can be defined as
$V = {V_0}\sin \left( {\omega t + {\theta _2}} \right)$..............…(ii)
Here, ${I_0}$ and ${V_0}$ are the peak values of the current and voltage respectively while I and V signify the values of current at some time t. ${\theta _1}$ and ${\theta _2}$ represent the phase angle for current and voltage respectively.
For an AC circuit, the average power dissipated is given as
$P = \dfrac{{{V_0}{I_0}}}{2}\cos \phi $
Here $\phi $ represents the phase difference between current and voltage given as
$\phi = {\theta _1} - {\theta _2}$

Complete step-by-step solution:
We are given an AC circuit and the expressions for the current and voltage in this circuit are given as
\[\begin{align}
& V = 200\sin \left( {319t - \dfrac{\pi }{6}} \right){\text{ volts}} \\
  &\Rightarrow I = 50\sin \left( {314t + \dfrac{\pi }{6}} \right){\text{ mA}} \\
\end{align} \]
Now we will compare these expressions to equations (i) and (ii), doing so we get
$\begin{align}
&{I_0} = 50mA = 50 \times {10^{ - 3}}A \\
&\Rightarrow {V_0} = 200V \\
&\Rightarrow {\theta _1} = \dfrac{\pi }{6} \\
&\Rightarrow {\theta _2} = - \dfrac{\pi }{6} \\
\end{align} $
The phase difference between the current and voltage is given as
$\phi = {\theta _1} - {\theta _2} = \dfrac{\pi }{6} - \left( { - \dfrac{\pi }{6}} \right) = \dfrac{{2\pi }}{6} = \dfrac{\pi }{3}$
The average power dissipated in a circuit is given as
$P = \dfrac{{{V_0}{I_0}}}{2}\cos \phi $
Now inserting all the known values, we get
$\begin{align}
 & P = \dfrac{{200 \times 50 \times {{10}^{ - 3}}}}{2}\cos \dfrac{\pi }{3} \\
 & = \dfrac{{200 \times 50 \times {{10}^{ - 3}}}}{2} \times \dfrac{1}{2} \\
 & = 2.5W \\
\end{align} $
This is the required answer. Hence, the correct answer is option A.

Note: Since in an AC circuit, the current and voltage keeps oscillating about the mean values, the circuit is characterized by its average power dissipated by the circuit. This value is just a mean value and the actual power dissipated by the circuit can be different at different instants of time.