For the circuit shown in the figure, the voltage and current expressions are
$$v\left( t \right) = {E_1}\sin \left( {\omega t} \right) + {E_3}\sin \left( {3\omega t} \right)$$ and
$$i\left( t \right) = {{\rm I}_1}\sin \left( {\omega t - {\varphi _1}} \right) + {{\rm I}_3}\sin \left( {3\omega t - {\varphi _3}} \right) + {{\rm I}_5}\sin \left( {5\omega t} \right)$$

The average power measured by the Wattmeter is

Consider the following statement:
(i) The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the current coil.
(ii) The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the voltage coil circuit.

A wattmeter is connected as shown in the fig. the wattmeter reads

A sampling wattmeter (that computes power from simultaneously sampled values of voltage and current) is used to measure the average power of a load. The peak to peak voltage of the square wave is $$10V$$ and the current is triangular wave of $$5A$$ $$p$$-$$p$$ as shown in the figure. The period is $$20$$$$ms$$. The reading in $$W$$ will be