A 20,F capacitor is charged to 5,V and isolated. It is then connected in parallel with an uncharged

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2. Electric Potential and Capacitance

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A $20\,F$ capacitor is charged to $5\,V$ and isolated. It is then connected in parallel with an uncharged $30\,F$ capacitor. The decrease in the energy of the system will be.......$J$

A

$25$

B

$200$

C

$125$

D

$150$

Solution

(d) $\Delta U = \frac{{{C_1}{C_2}}}{{2({C_1} + {C_2})}}{({V_1} – {V_2})^2}$$ = \frac{{20 \times 30}}{{2(20 + 30)}}{(5 – 0)^2} = 150\,J$

Standard 12

Physics

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