A 2 capacitor C _ 1 is first charged to a potential difference of...

A $2\muF$ capacitor ${C}_{1}$ is first charged to a potential difference of $10V$ using a battery. Then the battery is removed and the capacitor is connected to an uncharged capacitor ${C}_{2}$ of $8\muF$ . The charge in ${C}_{2}$ on equilibrium condition is $\muC$ . (Round off to the Nearest Integer)

A 2 capacitor C _ 1 is first charged to a potential difference of 10V using a battery. Then the battery is removed and

Correct answer

Step-by-step solution

$20 = \left({C}_{1} + {C}_{2}\right)V ⇒ V = 2volt .$

$⇒ {Q}_{2} = {C}_{2}V = 16\muC$

$= 16$

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