The potential difference $V$ across and the current flowing through an instrument are given by $V=5\cos \left( \omega t\, \right)\,\,V$ and $I=2\sin \left( \omega t \right)\,\,A$. The power dissipated in the instrument is-

Hint: This question involves the concept of A C circuit and analyzing phase angle between the alternating current and voltage passing through the instrument.
The steps involved to solve this question are-
1. Find the root mean square values of both current and voltage across the instrument.
2. Find the phase angle difference between both current and voltage.
3. Put the values in the formula of power dissipation in the instrument which is
  Power dissipated =${{V}_{RMS}}\times {{I}_{RMS}}\times \cos \varphi $

Complete step by step answer:
To find the power dissipated across the instrument we have to find the root mean square value.
Root mean values are the peak values or amplitude of a voltage or current function divided by $\sqrt{2}$.
${{V}_{RMS}}/{{I}_{RMS}}=\dfrac{{{V}_{PEAK}}/{{I}_{PEAK}}}{\sqrt{2}}$
According to question,
${{V}_{RMS}}=\dfrac{5}{\sqrt{2}}$
${{I}_{RMS}}=\dfrac{2}{\sqrt{2}}$
Now we have to find the phase difference,
We are given with $V=5\cos \left( \omega t\, \right)\,\,V$ and $I=2\sin \left( \omega t \right)\,\,A$,
As the voltage function is a cosine function ,lets convert it into sine function,
$V=5\cos \left( \omega t\, \right)\,\,V$=$5\sin \left( \omega t+\dfrac{\prod }{2} \right)V$
Now it can easily be seen that voltage across instrument leads current by $\varphi =\dfrac{\prod }{2}$ phase angle.putting the values in the formula for power dissipation,
${{V}_{RMS}}\times {{I}_{RMS}}\times \cos \varphi $
$\Rightarrow \dfrac{5}{\sqrt{2}}\times \dfrac{2}{\sqrt{2}}\times \cos \,{{90}^{o}}$
$\therefore \cos \,{{90}^{o}}=0$
Power dissipated =$0$

So the power dissipated across that instrument is $0$.

Note: In alternating current and voltages there are many types of values like root mean values ,root mean square values,peak values and average values over a period of time.
Resonance in an AC circuit is defined as a condition when the effective impedance of an Inductor ,capacitor and resistor circuit is reduced to resistance only and we achieve maximum current at the output.
Power dissipated most importantly depends on the phase angle between the alternating voltages and current.