The bridge method commonly used for finding mutual inductance is

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

A 220 V, 15 kW, 1000 rpm shunt motor with armature resistance of 0.25 Ω , has a rated line current of 68 A and a rated field current of 2.2 A. The change in field flux required to obtain a speed of 1600 rpm while drawing a line current of 52.8 A and a field current of 1.8 A is

A single phase 10 kVA, 50 Hz transformer with 1 kV primary winding draws 0.5 A and 55 W, at rated voltage and frequency, on no load. A second transformer has a core with all its linear dimensions $$\sqrt2$$ times the corresponding dimensions of the first transformer. The core material and lamination thickness are the same in both transformers. The primary windings of both the transformers have the same number of turns. If a rated voltage of 2 kV at 50 Hz is applied to the primary of the second transformer, then the no load current and power, respectively, are