JEE MAIN - Physics (2019 - 12th January Morning Slot - No. 9)
In the figure shown, after the switch 'S' is turned from position 'A' to position 'B', the energy dissipated in the circuit in terms of capacitance 'C' and total charge 'Q' is :
_12th_January_Morning_Slot_en_9_1.png)
$${1 \over 8}{{{Q^2}} \over C}$$
$${5 \over 8}{{{Q^2}} \over C}$$
$${3 \over 4}{{{Q^2}} \over C}$$
$${3 \over 8}{{{Q^2}} \over C}$$
Тайлбар
Vi = $${1 \over 2}$$CE2
Vf = $${{{{\left( {CE} \right)}^2}} \over {2 \times 4c}}$$ = $${1 \over 2}{{C{E^2}} \over 4}$$
$$\Delta $$E = $${1 \over 2}$$CE2 $$ \times $$ $${3 \over 4}$$ = $${3 \over 8}$$ CE2
Vf = $${{{{\left( {CE} \right)}^2}} \over {2 \times 4c}}$$ = $${1 \over 2}{{C{E^2}} \over 4}$$
$$\Delta $$E = $${1 \over 2}$$CE2 $$ \times $$ $${3 \over 4}$$ = $${3 \over 8}$$ CE2