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A capacitor of capacitance 2muF is charg...

A capacitor of capacitance `2muF` is charged to a potential difference of `5V`. Now, the charging battery is disconected and the capacitor is connected in parallel to a resistor of `5Omega` and another unknown resistor of resistance `R` as shown in figure. If the total heat produced in `5Omega` resistance is `10muJ` then the unknown resistance `R` is equal to

A

`10Omega`

B

`15Omega`

C

`10/3Omega`

D

`7.5Omega`

Text Solution

Verified by Experts

The correct Answer is:
C
Total heat produced `=1/2CV^2`
`=1/2(2muF)(5)^2`
`=2muJ`
Now this should distribute in inverse ratio of resistors, as they are in parallel.
`:. H_(5Omega)/H_R=R/5`
or `H_(5Omega)=(R/(R+5))` (total heat)
`or 10=(R/(R+5))(25)`
Solving this equation, we get
`R=(10/3)Omega`
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DC PANDEY-CAPACITORS-Exercise
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