Physics, asked by nalin23aaditya, 10 months ago

A capacitor of 2 uF is charged as shown in the
diagram. When the switch S is turned to position
2, the percentage of its stored energy dissipated is

(1) 80%
(3) 20%
(2) 0%
(4) 75%

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Answers

Answered by Anonymous
86

Answer

\sf{\green{Option\:(1)}\:80\%}

\rule{150}2

Energy stored in parallel plate capacitor

Consider a parallel plate capacitor is charged to potential (V), then small amount of work done to add another charge (dq) is

\sf{dw\: = \:Vdq}

\sf{dw\:=\:\dfrac{q}{C}dq}

\displaystyle{\sf{\int\limits_{0}^{w}dw\:=\:\dfrac{1}{C}\int\limits_{0}^{q}dq}}

\sf{w \:  =  \:  \dfrac{1}{C} \bigg( \dfrac{ {q}^{2} }{2} \bigg)_{0}^{q} }

\sf{w\:=\:\dfrac{q^2}{2C}}

This work done is stored as energy stored in parallel plate capacitor.

\sf{U\:=\:\dfrac{q^2}{2C}}

\sf{U\:=\:\dfrac{C^2V^2}{2C}\:=\:\dfrac{1}{2}CV^2} ...(1)

\sf{U\:=\:\dfrac{1}{2}(CV)V\:=\:\dfrac{1}{2}qV}

According to question,

We have given, \sf{C\: = \:2\mu\:F\:=\:2\:\times\:10^{-6}}

Substitute value of C in (1)

=> \sf{U\:=\:\dfrac{1}{2}(2\:\times\:10^{-6}V^2)}

=> \sf{U\:=\:10^{-6}V^2\:J(joule)}

Now,

Capacitor :

It is a device which is used to store electrical charge and release it at once.

\sf{q\:\alpha\:V}

\sf{q\:=\:CV}

(q = 2 × 10^-6 × V)

\sf{V\:=\:\dfrac{q}{C}}

=> \sf{V\:=\:\dfrac{2\:\times\:10^{-6}\:\times\:V}{(2\:+\:8)\:\times\:10^{-6}}}

=> \sf{V\:=\:\dfrac{2V}{10}}

=> \sf{V\:=\:\dfrac{V}{5}\:Volt}

Now, to find the final energy stored in the capacitor. Substitute value of V = V/5 Volt and C = 10 × 10^-6 = (2+8) 10^-6 in equation (1)

=> \sf{U'\:=\:\dfrac{1}{2}(10\:\times\:10^{-6})\bigg(\dfrac{V}{5}\bigg)^2}

=> \sf{U'\:=\:5\:\times\:10^{-6}\:\dfrac{V^2}{25}}

=> \sf{U'\:=\:10^{-6}\:\dfrac{V^2}{5}\:J}

Now, loss of energy in percentage = \sf{\dfrac{U\:-\:U'}{U}\:\times\:100\%}

=> \sf{\dfrac{V^2\:\times\:10^{-6}\:-\:\frac{V^2}{5}\:\times\:10^{-6}}{V^2\:\times\:10^{-6}}\times\:100\%}

=> \sf{\dfrac{\bigg(V^2\:-\:\dfrac{V^2}{5}\bigg)10^{-6}}{V^2\:\times\:10^{-6}}\:\times\:100\%}

=> \sf{\dfrac{\bigg(\dfrac{4V^2}{5} \bigg)}{V^2}\:\times\:100\%}

=> \sf{0.8\:\times\:100\%}

=> \sf{80\%}

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