Consider the circuit shown. Assume that the diode $D$ is ideal. The supply voltage $v_s=325 \sin (2 \pi 50) \vee, L=500 \mu \mathrm{H}$, and $R=10 \Omega$. The peak diode current (in amperes) is $\_\_\_\_$ . (Round off to one decimal place)

A single-phase full-bridge diode rectifier feeds a resistive load of 50 $$\Omega$$ from a 200 V, 50 Hz single phase AC supply. If the diodes are ideal, then the active power, in watts, drawn by the load is _________. (round off to nearest integer).

The voltage at the input of an AC-DC rectifier is given by $$v(t) = 230\sqrt 2 \sin \omega t$$ where $$\omega = 2\pi \times 50$$ rad/s. The input current drawn by the rectifier is given by

$$i(t) = 10\sin \left( {\omega t - {\pi \over 3}} \right) + 4\sin \left( {3\omega t - {\pi \over 6}} \right) + 3\sin \left( {5\omega t - {\pi \over 3}} \right)$$

The input power factor, (rounded off to two decimal places), is _________ lag.