A capacitor is charged with a battery and the energy stored is $U$ . After disconnecting battery another capacitor of the same capacity is connected in parallel with it. Then the energy stored in each capacitor is :
$\frac{U}{2}$
$\frac{U}{4}$
$4 U$
$2 U$

Question

A capacitor is charged with a battery and the energy stored is $U$ . After disconnecting battery another capacitor of the same capacity is connected in parallel with it. Then the energy stored in each capacitor is :
$\frac{U}{2}$
$\frac{U}{4}$
$4 U$
$2 U$

Toppr Admin · Accepted Answer

Energy stored-xA0-U-q22C-xA0-After connecting with another capacitorVcommon-q1-q2C1-C2-q-0C-C-q2C -xA0-x2235-xA0-Energy on each capacitor-xA0-12CV2Common-12-q2C-2-U4-

A capacitor is charged with a battery and the energy stored is U. After disconnecting battery another capacitor of the same capacity is connected in parallel with it. Then the energy stored in each capacitor is :U2U44U2U

Energy stored (U)=q22C After connecting with another capacitorVcommon=q1+q2C1+C2=q+0C+C=q2C ∵ Energy on each capacitor =12CV2Common=12(q2C)2=U4.

A capacitor is charged with a battery and the energy stored is $U$ . After disconnecting battery another capacitor of the same capacity is connected in parallel with it. Then the energy stored in each capacitor is :
$\frac{U}{2}$
$\frac{U}{4}$
$4 U$
$2 U$

Energy stored $(U) = \frac{q^{2}}{2 C}$ After connecting with another capacitor

$V_{c o m m o n} = \frac{q_{1} + q_{2}}{C_{1} + C_{2}} = \frac{q + 0}{C + C} = \frac{q}{2 C}$

$∵$ Energy on each capacitor

$= \frac{1}{2} C V_{C o m m o n}^{2} = \frac{1}{2} {(\frac{q}{2 C})}^{2} = \frac{U}{4} .$