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.......$J$
Medium
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(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$
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