Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsMiscellaneousEasy2 minPYQ_2020

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A $2$ $μ F$ capacitor is charged as shown in the figure. The percentage of its stored energy dissipated as heat after the switch $S$ is turned to position $2$ is

$D:\Session 2018-2019\Online Test Paper\Online CBT Series-2018\Phase-II\Test-11_(07-03-2019)\Images\q18.jpg$

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Answer:

Solution:

$E (i n i t i a l) = \frac{Q_{0}^{2}}{2 C} = \frac{1}{2} C V_{0}^{2} = V_{0}^{2}$
Final energy $E (f) = \frac{Q_{0}^{2}}{2 C_{e f f}}$
$E (f) = \frac{1}{2} \times \frac{2 v_{0}}{10} \times 2 v_{0} = \frac{v_{0}^{2}}{5}$
$Δ E (l o s s) = \frac{4 v_{0}^{2}}{5},$
$%$ loss = $\frac{4}{5} \frac{v_{0}^{2}}{v_{0}^{2}} \times 100$
$Δ E (l o s s) % = 80 %$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:ElectrostaticsSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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