Question

- Consider the situation shown. The switch S is open for a long time and then closed Then :

| Column I| | Column II
---|---|---|---
(A)| Charge flown through battery when S is closed| (p)| $\dfrac{{C{E^2}}}{2}$
(B)| Work done by battery | (q)| $\dfrac{{CE}}{2}$
(C)| Change in energy stored in capacitor| (r)| $\dfrac{{C{E^2}}}{4}$
(D)| Heat developed in the system| |

Answer 1

If the switch is closed then one capacitor becomes in series combination and other is short-circuited means no charge is stored as both the terminals are connected to the same side of the battery. And also heat developed in the system is equal to the change in energy.

Complete step by step solution:
Initially, both the capacitors are in series combination ${C_{eq}} = \dfrac{{C \times C}}{{2C}}$
$\Rightarrow {C_{eq}} = \dfrac{{C \times C}}{{2C}}$
Simplifying the equation we get ,
$\Rightarrow {C_{eq}} = \dfrac{C}{2}$.
If the switch is closed then one capacitor becomes in series combination and other is short-circuited means no charge is stored as both the terminals are connected to the same side of the battery.
In that case ${C_{final}} = C + 0$
$\Rightarrow {C_{final}} = C + 0$
$\Rightarrow {C_{final}} = C$
Initial capacitance was ${C_{initial}} = \dfrac{C}{2}$ & Final capacitance is ${C_{final}} = C$.
Initially the switch is open and ${C_{initial}} = \dfrac{C}{2}$, if the switch is closed then first capacitor discharges
Charge flown through battery when switch ( S ) is closed is
$\Rightarrow Q = ({C_{initial}} - {C_{disch\arg ing}}) \times E$
Putting the values we get ;
$\Rightarrow Q = \left( {\dfrac{C}{2} - 0} \right) \times E$
Simplifying it we get,
$\Rightarrow Q = \dfrac{{CE}}{2}$.
Work done by battery ( W )
$\Rightarrow W = \dfrac{C}{2}{E^2}$
$\Rightarrow W = \dfrac{{C{E^2}}}{2}$.
Change in energy stored in capacitor
$\Rightarrow {C_{initial}} = \dfrac{C}{2}$
$\Rightarrow {C_{final}} = C$
Now, we have to find the energy stored initially, so for this
$\Rightarrow {E_{initial}} = \dfrac{1}{2} \times \dfrac{C}{2} \times {E^2}$
Simplifying the equation we get ,
$\Rightarrow {E_{initial}} = \dfrac{{C{E^2}}}{4}$
Now, we have to find the energy stored finally, so for this
$\Rightarrow {E_{final}} = \dfrac{1}{2} \times C \times {E^2}$
$\Rightarrow {E_{final}} = \dfrac{{C{E^2}}}{2}$.
Change in energy stored in the capacitor is ${E_{final}} - {E_{initial}}$.
Putting the values we get with ,
$\Rightarrow {E_{final}} - {E_{initial}} = \dfrac{{C{E^2}}}{2} - \dfrac{{C{E^2}}}{4}$
Simplifying the equation we get,
$\Rightarrow {E_{final}} - {E_{initial}} = \dfrac{{C{E^2}}}{4}$……… eq (1)

Heat developed in the system $H = \Delta E$.
$\Rightarrow H = {E_{final}} - {E_{initial}}$
Using the result obtained from eq (1)
$\Rightarrow H = {E_{final}} - {E_{initial}} = \dfrac{{C{E^2}}}{4}$
$\Rightarrow H = \dfrac{{C{E^2}}}{4}$.

Hence the correct matches are A-q, B-p, C-r and D-r.

Note: The energy of an uncharged capacitor is zero. When a capacitor is charged by battery, both the plates received charge equal in magnitude, no matter the size of plates are identical or not because the charge distribution on the plates of a capacitor is in accordance with charge conservation principle.