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A capacitor of 2\muF is charged as shown in the diagram. When the switch S is turn to position 2, the percentage of its stored energy dissipated is:

 

  • Option 1)

       0%

  • Option 2)

    20%

  • Option 3)

    75%

  • Option 4)

    80%

 

Answers (1)

best_answer

As we discussed

Loss of Energy -

Delta U=U_{i}-U_{f}=frac{C_{1}C_{2}}{2left ( C_{1} +C_{2} ight )}left ( V_{1}-V_{2} ight )^{2}

-

 

 

Energy Stored -

U=frac{1}{2}CV^{2}=frac{1}{2}QV=frac{Q^{2}}{2C}

-

 

 when s and 1 are connected

the 2\mu f capacitor gets charged

u_{i}=\frac{1}{2}cv^{2}=\frac{1}{2}\times 2\times v^{2}=v^{2}

when S and 2 are connected

the 8\mu f capaciter also gets charged

the energy loss \Delta u

\Delta u=\frac{1}{2}\frac{c_{1}c_{2}}{c_{1}+c_{2}}\left ( v_{1}-v_{2} \right )^{2}

c_{1}=2\mu f

c_{2}=8\mu f

v_{1}=v \: v_{2}=0

\therefore \Delta u=\frac{1}{2}\times \frac{2\times 8}{2+8}\left ( v-0 \right )^{2} = \frac{4}{5}^{2}

Percentage of energy dissipated=\frac{\Delta v}{v_{i}}\times 100=\frac{\frac{4}{5}v^{2}}{v^{2}} \times 100

=80%


Option 1)

   0%

This is incorrect option

Option 2)

20%

This is incorrect option

Option 3)

75%

This is incorrect option

Option 4)

80%

This is correct option

Posted by

divya.saini

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