Saved Bookmarks
| 1. |
Find total energy stored in capacitors given in the circuit. |
|
Answer» Solution :Capacitors `C_(4) and C_(5)` are in series, their EQUIVALENT capacitance `C.=(C_(4)C_(5))/(C_(4)+C_(5))=(2xx2)/(2+2)=1muF` C. and `C_(2)` are in parallel having equivalent capacitance `C..=C_(2)+C.=1+1=2muF` C.. and `C_(3)` are in series, their equivalent capacitance `C_(0)=(C..C_(3))/(C..+C_(3))=(2xx2)/(2+2)=1muF` `therefore` Total energy stored in the capacitors in the given network `U=(1)/(2)CV^(2)=(1)/(2)xx2xx10^(-6)xx(6)^(2)=3.6xx10^(-5)J`. |
|