A 5 μF capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series

Question

A 5 $\mu$F capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5 $\mu$F capacitor. If the energy change during the charge redistribution is $X\over100$ J then value of X to the nearest integer is : 

Numerical Value Type

Comments

Electrostatics, electrostatic potential and capacitance

Answer 1

Ans. 4

NTA answer is 36 but JeeNeetQnA answer is 4

Sol. C₁ = 5 $\mu$F    V₁ = 220 Volt

C₂ = 2.5 $\mu$F    V₂ = 0

Heat loss: $\Delta H = U_i – U_f = {1\over2}{C_1C_2\over C_1+C_2}(V_1-V_2)^2$

$={1\over2}\times{5\times2.5\over(5+2.5)}(220-0)^2\mu J$

$={5\over2\times3}×22×22×100×10^{-6}J$

$={5×11×22\over3}×10^{-4}J={55×22\over3}×10^{-4}J$

$={1210\over3}×10^{-4}J={121\over3}×10^{-3}J$

$=4×10^{-2}J$

According to questions

${x\over100}=4×10^{-2}$

So, x = 4

Note : But given answer by JEE Main x = 36

Answer 2

Ans: 4

Explanation: