Q. The output in the circuit of figure is taken across a capacitor. It is as shown in figure
A) [image:14_image199.png]
B) [image:14_image200.png]
C) [image:14_image201.png]
D) [image:14_image202.png]
Explanation
(c) As $RC$ time constant of the capacitor is quite large $(\tau = RC = 10 \times 10^{3} \times 10 \times 10^{- 6} = 0.1\ s)$, it will not discharge appreciably. Hence voltage remains nearly constant
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.