Question

Two capacitors, 3 $\mu F$ and 4 $\mu F$, are individually charged across a 6 V battery. After being disconnected from the battery, they are connected together with the negative plate of one attached to the positive plate of the other. What is the final total energy stored?

• A. $1.26\times {10}^{-4}$ J
• B. $2.57\times {10}^{-4}$ J
• C. $1.26\times {10}^{-6}$ J
• D. $2.57\times {10}^{-6}$ J

Answer 1

Answer: (D)

$2.57\times {10}^{-6}$ J

As $Q=CV$

$\therefore$ Initially charge on each capacitor is

$Q_1=C_1{V}_1=\left(3\mu F\right)\left(6V\right)=18\mu C$ and $Q_2=C_2{V}_2=\left(4\mu F\right)\left(6V\right)=24\mu C$

When two capacitors are joined to each other such that negative plate of one is attached with the positive plate of the other. The charges $Q_1$ and $Q_2$ are redistributed till they attain the common potential which is given by

Common potential $V=\frac{Totalcharge}{Totalcapacitiance}$

$=\frac{24\mu C-18\mu C}{3\mu F+4\mu F}=\frac{6}{7}V$

Final energy stored

$U_f=\frac{1}{2}(C_1+C_2)V^2=\frac{1}{2}[3\times {10}^{-6}+4\times {10}^{-6}]\times {\left(\frac{6}{7}\right)}^2=\frac{1}{2}\times 7\times {10}^{-6}\times \frac{36}{49}=2.57\times {10}^{-6}J$