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Question:

A $10 \mu \mathrm{F}$ capacitor is fully charged to a potential difference of $50 \mathrm{~V}$. After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes $20 \mathrm{~V}$. The capacitance of the second capacitor is :

  1. (1) $15 \mu \mathrm{F}$

  2. (2) $30 \mu \mathrm{F}$

  3. (3) $20 \mu \mathrm{F}$

  4. (4) $10 \mu \mathrm{F}$


Correct Option: 1

Solution:

(1) Given,

Capacitance of capacitor, $C_{1}=10 \mu \mathrm{F}$

Potential difference before removing the source voltage, $V_{1}=50 \mathrm{~V}$

If $C_{2}$ be the capacitance of uncharged capacitor, then common potential is

$V=\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}$

$\Rightarrow 20=\frac{10 \times 50+0}{20+C} \Rightarrow C=15 \mu F$

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