Question

A condenser of certain capacity is charged to a potential V and stores some energy. A second condenser of twice of the capacity is to be stored half the energy of the first .
Find upto what potential one must be charged

pls give relevant answer

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Answer 1

Answer:

It should be charged upto half of the first, that is, V/2

Explanation:

We know that energy of a capacitor in terms of capacitance and potential is,

$\boxed {E = \dfrac{1}{2} CV^2}$

Given, first capacitor has charged to a potential V and stores some energy. [Let us assume this first capacitor has capacitance C]

this energy is equal to [E₁ means energy of first capacitor]

${E_1 = \dfrac{1}{2} CV^2}$

Now, it is given that, A second condenser of twice of the capacity is to be stored half the energy of the first.

It means, C₂ = 2C and E₂ = 1/2E₁

Writing the formula for E₂, we get,

Let the potential of second capacitor be V'

$E_2 = \dfrac{1}{2} (2C)V'^2$

$: \implies \dfrac{1}{2} E_1 = \dfrac{1}{2} (2C)V'^2$

$: \implies \bcancel{\dfrac{1}{2}} E_1 = \bcancel{\dfrac{1}{2} }(2C)V'^2$

$: \implies E_1 = (2C)V'^2$

$: \implies \dfrac{1}{2} CV^2= (2C)V'^2$

$: \implies \dfrac{1}{4} V^2= V'^2$

$: \implies V' = \dfrac{1}{2} V$

$\therefore \mathbf{V' = \dfrac{V}{2} }$

Hope it helps!!