Question

The capacitor C is initially without charge. X is now joined to Y for a long time, during which $H_1$ heat is produced in the resistance R. X is now joined to Z for a long time, during which $H_2$ heat is produced in R

• A. $H_1=H_2$
• B. $H_1=\frac{1}{2}{H}_2$
• C. $H_1=2H_2$
• D. the maximum energy stored in C at any time is $H_1$

Answer 1

Answer: (A), (D)

The correct options are
A $H_1=H_2$
D the maximum energy stored in C at any time is $H_1$

Assuming that the source is of V Voltage.

Energy dissipated by the Source $\int IVdt=Q\times V$

and $Q=CV$

therefore Total Energy from Source $=C{V}^2$

Maximum energy that can be stored in a Capacitor $=\frac{1}{2}\times C{V}^2$

Hence $H_1=C{V}^2-\frac{1}{2}\times C{V}^2=\frac{1}{2}\times C{V}^2=$ Max Energy in Capacitor C

and $H_2=\frac{1}{2}\times C{V}^2$ = Energy Drain from the Capacitor

Hence Answer is A & D