Question

A 5µF 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 µF capacitor If the energy change during the charge redistribution is `"X"/100`J then value of X to the 100 nearest integer is ______.

पर्याय

• 2

• 3

• 4

• 5

Answer 1

A 5µF 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 µF capacitor If the energy change during the charge redistribution is `"X"/100`J then value of X to the 100 nearest integer is 4.

Explanation:

Given: Capacitance of capacitor - 1 is C₁ = 5 µF, voltage across capacitor - 1 is V₁ = 220 V capacitance of capacitor - 2 is C₂ = 2.5 µF, voltage across capacitor - 2 is V₂ = 0V, the two capacitors are then connected in series, the change in energy of the system during the charge redistribution is

ΔE = `"X"/100` J.

To find: The value of X.

Energy lost during the charge redistribution:

ΔE = `1/2 ("C"_1"C"_2)/(2"C"_1+"C"_2)xx("V"_1 -"V"_2)^2`

ΔE = `4/100`J

But ΔE = `"X"/100`J

So, X = 4