The output in the circuit of figure is taken across a capacitor. It is as shown in figure

Options

A.
B.
C.
D.

Answer

C

Explanation

For `+` ve half `to` D `to` ON
For `-ve` half `to` D `to` OFF
In `+` ve half cycle, capacitor charges but in `-ve` half it discharges very slowly because
Time constant `=tauR=10xx10^(3)xx10xx10^(-6)=0.1"sec"`