Question

A capacitor $10\mu F$ charged to 50 V is joined to another uncharged $50\mu C$ capacitor. Find the loss in energy.

• A. $1.04\times {10}^{-2}J$
• B. $4.01\times {10}^{-2}J$
• C. $6.25\times {10}^{-4}J$
• D. $1.64\times {10}^{-4}J$

Answer 1

The correct option is A $1.04\times {10}^{-2}J$

When two capacitors are joined, there is always a loss of energy, which is given by:

$\Delta U=\frac{C_1{C}_2(V_1-V_2{)}^2}{2(C_1+C_2)}$,

given : $C_1=10\mu F=10\times {10}^{-6}F,C_2=50\mu F=50\times {10}^{-6}F,V_1=50V,V_2=0V$,

therefore,

$\Delta U=\frac{10\times {10}^{-6}\times 50\times {10}^{-6}\times (50-0{)}^2}{2(10\times {10}^{-6}+50\times {10}^{-6})}$,

or $\Delta U=\frac{125\times {10}^{-2}}{120}=1.04\times {10}^{-2}J$