Question 54: Electrostats & Capacitance

Question

A capacitor of capacity $$2 ~\mu \mathrm{F}$$ is charged upto a potential $$14 \mathrm{~V}$$ and then connected in parallel to an uncharged capacitor of capacity $$5 ~\mu \mathrm{F}$$. The final potential difference across each capacitor will be

Accepted Answer

When two capacitors are connected then their common potential difference is
$$V=\frac{C_1 V_1+C_2 V_2}{C_1+C_2}$$
Given, $$C_1=2 \mu \mathrm{F}, V_1=14 \mathrm{~V}, C_2=5 \mu \mathrm{F}, V_2=0~V$$
$$	herefore \quad V=\frac{2 	imes 14+5 	imes 0}{2+5}=4 V$$

Answer

6 V

Answer

4 V

Answer

8 V

Answer

14 V

Electrostats & Capacitance

A capacitor of capacity $2 ~\mu \mathrm{F} is charged upto a potential 14 \mathrm{~V} and then connected in parallel to an uncharged capacitor of capacity 5 ~\mu \mathrm{F}$. The final potential difference across each capacitor will be

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