A capacitor of capacitance $2\,\mu F$ is charged to $40\,\,{\rm{V}}$ and another capacitor of capacitance $4\,\mu F$ is changed to $20\,\,{\rm{V}}.$ If the capacitors are connected together in same polarity, then the energy lost in reorganisation of charge will be

JEE PHYSICS MCQ: ELECTRIC POTENTIAL AND CAPACITANCE

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Application +4 | -1

A capacitor of capacitance

2\,\mu F

is charged to

40\,\,\mathrm{V}

and another capacitor of capacitance

4\,\mu F

is changed to

20\,\,\mathrm{V}.

If the capacitors are connected together in same polarity, then the energy lost in reorganisation of charge will be

View answer and explanation

Correct option: B

266.67\,\,\mu J

Option B (

266.67 \,\,\mu J

) is correct because it is the energy lost calculated using the formula for energy loss during capacitor connection.

Practice All Chapter MCQs →

x

\mu \mathrm{m}

x

V(x)=\frac{20}{x^2-5} \mathrm{V}

x=5 \mu \mathrm{m}

R

Q

3\mu F,\,

16\mu F

E

E